第 3 章 恒等式

Chapter 3 Identities

1. 基本三角恒等式

Basic Trigonometric Identities

到目前为止，我们知道了三角函数之间的一些关系。例如，我们知道以下互逆关系：

1. 当 \(\sin\;\theta \ne 0\) 时， \(\csc\;\theta ~=~ \dfrac{1}{\sin\;\theta}\qquad\)

2. 当 \(\cos\;\theta \ne 0\) 时， \(\sec\;\theta ~=~ \dfrac{1}{\cos\;\theta}\qquad\)

3. 当 \(\tan\;\theta\) 有定义且不为 0 时， \(\cot\;\theta ~=~ \dfrac{1}{\tan\;\theta}\qquad\)

4. 当 \(\sin\;\theta ~=~ \dfrac{1}{\csc\;\theta}\qquad\) \(\csc\;\theta\) 有定义且不为 0

5. \(\cos\;\theta ~=~ \dfrac{1}{\sec\;\theta}\qquad\)，当 \(\sec\;\theta\) 有定义且不为 0

6. \(\tan\;\theta ~=~ \dfrac{1}{\cot\;\theta}\qquad\)，当 \(\cot\;\theta\) 有定义且不为 0

请注意，对于等式两边均有定义的 所有 角 \(\theta\)，这些等式均成立。这类等式称为 恒等式(identities) ，本节我们将讨论几个 三角恒等式 ，即涉及三角函数的恒等式。这些恒等式常用于简化复杂的表达式或方程。例如，最有用的三角恒等式之一如下：

(1)\[\tan\;\theta ~=~ \frac{\sin\;\theta}{\cos\;\theta} \qquad \text{当 } \cos\;\theta \ne 0\]

为了证明这个恒等式，在 \(\theta\) 的端点处取一个点 $(x,y)$，该点距离原点 $r > 0$，并假设 \(\cos\;\theta \ne 0\)。则 \(x \ne 0\) （因为 \(\cos\;\theta = \frac{x}{r}\) ），所以根据定义

\[\frac{\sin\;\theta}{\cos\;\theta} ~=~ \dfrac{~\dfrac{y}{r}~}{~\dfrac{x}{r}~} ~=~ \frac{y}{x} ~=~ \tan\;\theta ~。\]

注意，我们是如何通过展开其中一边（ \(\frac{\sin\;\theta}{\cos\;\theta}\) ）来证明恒等式的，直到得到一个等于另一边的表达式（\(\tan\;\theta\)）。这可能是证明恒等式最常用的技巧。对上述恒等式取倒数可得：

(2)\[\cot\;\theta ~=~ \frac{\cos\;\theta}{\sin\;\theta} \qquad \text{当 } \sin\;\theta \ne 0\]

图 3.1.1

现在我们将推导一个最重要的三角恒等式。设 \(\theta\) 为任意角，其端点 $(x,y)$ 到原点的距离为 $r>0$。根据勾股定理，\(r^2 = x^2 + y^2\) （因此 \(r=\sqrt{x^2 + y^2}\)）。

例如，如果 \(\theta\) 位于 QIII 中，如图 3.1.1 所示，则由参考角构成的直角三角形的两条边的长度分别为 \(|x|\) 和 \(|y|\) （我们使用绝对值，因为 $x$ 和 $y$ 在 QIII 中为负数）。如果 \(\theta\) 位于其他象限或任一轴上，则同样的道理也成立。因此，

\[r^2 ~=~ |x|^2 ~+~ |y|^2 ~=~ x^2 ~+~ y^2 ~，\]

因此，等式两边同时除以 $r^2$（由于 $r>0$，我们可以这样做）可得：

\[\frac{r^2}{r^2} ~=~ \frac{x^2 ~+~ y^2}{r^2} ~=~ \frac{x^2}{r^2} ~+~ \frac{y^2}{r^2} ~=~ \left(\frac{x}{r}\right)^2 ~+~ \left(\frac{y}{r}\right)^2 ~。\]

由于 \(\frac{r^2}{r^2} = 1$，$\frac{x}{r} = \cos\;\theta\)，且 \(\frac{y}{r} = \sin\;\theta\)，我们可以将其重写为：

(3)\[\cos^2 \;\theta ~+~ \sin^2 \;\theta ~=~ 1\]

你可以将其视为勾股定理的三角函数变体。注意，我们使用 \(\sin^2 \;\theta\) 表示 \((\sin\;\theta)^2\)，余弦和其他三角函数也是如此。我们将对除 2 之外的其他幂使用相同的符号。

从上述恒等式我们可以推导出更多的恒等式。例如：

(4)\[\sin^2 \;\theta ~=~ 1 ~-~ \cos^2 \;\theta\]

(5)\[\cos^2 \;\theta ~=~ 1 ~-~ \sin^2 \;\theta\]

由此我们得到（开平方后）：

(6)\[\sin\;\theta ~=~ \pm\,\sqrt{1 ~-~ \cos^2 \;\theta}\]

(7)\[\cos\;\theta ~=~ \pm\,\sqrt{1 ~-~ \sin^2 \;\theta}\]

此外，从不等式 \(0 \le \sin^2 \;\theta = 1 ~-~ \cos^2 \;\theta \le 1\) 和 \(0 \le \cos^2 \;\theta = 1 ~-~ \sin^2 \;\theta \le 1\) 取平方根可得出正弦和余弦的以下界限：

(8)\[-1 ~ \le ~ \sin\;\theta ~ \le ~ 1\]

(9)\[-1 ~ \le ~ \cos\;\theta ~ \le ~ 1\]

上述不等式并非恒等式（因为它们并非方程），但它们有助于检验计算结果。回想一下，我们根据 1.4 节 中正弦和余弦的定义推导出了这些不等式。

在公式 (3) 中，恒等式两边同时除以 \(\cos^2 \;\theta\) 可得

\[\frac{\cos^2 \;\theta}{\cos^2 \;\theta} ~+~ \frac{\sin^2 \;\theta}{\cos^2 \;\theta} ~=~ \frac{1}{\cos^2 \;\theta} ~~,\]

因此，由于 \(\tan\;\theta = \frac{\sin\;\theta}{\cos\;\theta}\) 且 \(\sec\;\theta = \frac{1}{\cos\;\theta}\)，我们得到：

(10)\[1 ~+~ \tan^2 \;\theta ~=~ \sec^2 \;\theta\]

同样，将公式 (3) 两边除以 \(\sin^2 \;\theta\) 可得：

\[\frac{\cos^2 \;\theta}{\sin^2 \;\theta} ~+~ \frac{\sin^2 \;\theta}{\sin^2 \;\theta} ~=~ \frac{1}{\sin^2 \;\theta} ~~，\]

因此由于 \(\cot\;\theta = \frac{\cos\;\theta}{\sin\;\theta}\) 和 \(\csc\;\theta = \frac{1}{\sin\;\theta}\)，我们得到：

(11)\[\cot^2 \;\theta ~+~ 1 ~=~ \csc^2 \;\theta\]

So far we know a few relations between the trigonometric functions. For example, we know the reciprocal relations:

1. \(\csc\;\theta ~=~ \dfrac{1}{\sin\;\theta}\qquad\) when \(\sin\;\theta \ne 0\)

2. \(\sec\;\theta ~=~ \dfrac{1}{\cos\;\theta}\qquad\) when \(\cos\;\theta \ne 0\)

3. \(\cot\;\theta ~=~ \dfrac{1}{\tan\;\theta}\qquad\) when \(\tan\;\theta\) is defined and not $0$

4. \(\sin\;\theta ~=~ \dfrac{1}{\csc\;\theta}\qquad\) when \(\csc\;\theta\) is defined and not $0$

5. \(\cos\;\theta ~=~ \dfrac{1}{\sec\;\theta}\qquad\) when \(\sec\;\theta\) is defined and not $0$

6. \(\tan\;\theta ~=~ \dfrac{1}{\cot\;\theta}\qquad\) when \(\cot\;\theta\) is defined and not $0$

Notice that each of these equations is true for all angles \(\theta\) for which both sides of the equation are defined. Such equations are called identities, and in this section we will discuss several trigonometric identities, i.e. identities involving the trigonometric functions. These identities are often used to simplify complicated expressions or equations. For example, one of the most useful trigonometric identities is the following:

\[\tan\;\theta ~=~ \frac{\sin\;\theta}{\cos\;\theta} \qquad \text{when } \cos\;\theta \ne 0\]

To prove this identity, pick a point $(x,y)$ on the terminal side of \(\theta\) a distance $r >0$ from the origin, and suppose that \(\cos\;\theta \ne 0\). Then \(x \ne 0\) (since \(\cos\;\theta = \frac{x}{r}\)), so by definition

\[\frac{\sin\;\theta}{\cos\;\theta} ~=~ \dfrac{~\dfrac{y}{r}~}{~\dfrac{x}{r}~} ~=~ \frac{y}{x} ~=~ \tan\;\theta ~.\]

Note how we proved the identity by expanding one of its sides (\(\frac{\sin\;\theta}{\cos\;\theta}\)) until we got an expression that was equal to the other side (\(\tan\;\theta\)). This is probably the most common technique for proving identities. Taking reciprocals in the above identity gives:

\[\cot\;\theta ~=~ \frac{\cos\;\theta}{\sin\;\theta} \qquad \text{when } \sin\;\theta \ne 0\]

Figure 3.1.1

We will now derive one of the most important trigonometric identities. Let \(\theta\) be any angle with a point $(x,y)$ on its terminal side a distance $r>0$ from the origin. By the Pythagorean Theorem, \(r^2 = x^2 + y^2\) (and hence \(r=\sqrt{x^2 + y^2}\)).

For example, if \(\theta\) is in QIII as in Figure 3.1.1, then the legs of the right triangle formed by the reference angle have lengths \(|x|\) and \(|y|\) (we use absolute values because $x$ and $y$ are negative in QIII). The same argument holds if \(\theta\) is in the other quadrants or on either axis. Thus,

\[r^2 ~=~ |x|^2 ~+~ |y|^2 ~=~ x^2 ~+~ y^2 ~,\]

so dividing both sides of the equation by $r^2$ (which we can do since $r>0$) gives

\[\frac{r^2}{r^2} ~=~ \frac{x^2 ~+~ y^2}{r^2} ~=~ \frac{x^2}{r^2} ~+~ \frac{y^2}{r^2} ~=~ \left(\frac{x}{r}\right)^2 ~+~ \left(\frac{y}{r}\right)^2 ~.\]

Since \(\frac{r^2}{r^2} = 1$, $\frac{x}{r} = \cos\;\theta\), and \(\frac{y}{r} = \sin\;\theta\), we can rewrite this as:

\[\cos^2 \;\theta ~+~ \sin^2 \;\theta ~=~ 1\]

You can think of this as sort of a trigonometric variant of the Pythagorean Theorem. Note that we use the notation \(\sin^2 \;\theta\) to mean \((\sin\;\theta)^2\), likewise for cosine and the other trigonometric functions. We will use the same notation for other powers besides $2$.

From the above identity we can derive more identities. For example:

\[\sin^2 \;\theta ~=~ 1 ~-~ \cos^2 \;\theta\]

\[\cos^2 \;\theta ~=~ 1 ~-~ \sin^2 \;\theta\]

from which we get (after taking square roots):

\[\sin\;\theta ~=~ \pm\,\sqrt{1 ~-~ \cos^2 \;\theta}\]

\[\cos\;\theta ~=~ \pm\,\sqrt{1 ~-~ \sin^2 \;\theta}\]

Also, from the inequalities \(0 \le \sin^2 \;\theta = 1 ~-~ \cos^2 \;\theta \le 1\) and \(0 \le \cos^2 \;\theta = 1 ~-~ \sin^2 \;\theta \le 1\), taking square roots gives us the following bounds on sine and cosine:

\[-1 ~ \le ~ \sin\;\theta ~ \le ~ 1\]

\[-1 ~ \le ~ \cos\;\theta ~ \le ~ 1\]

The above inequalities are not identities (since they are not equations), but they provide useful checks on calculations. Recall that we derived those inequalities from the definitions of sine and cosine in Section 1.4.

In formula (3), dividing both sides of the identity by \(\cos^2 \;\theta\) gives

\[\frac{\cos^2 \;\theta}{\cos^2 \;\theta} ~+~ \frac{\sin^2 \;\theta}{\cos^2 \;\theta} ~=~ \frac{1}{\cos^2 \;\theta} ~~,\]

so since \(\tan\;\theta = \frac{\sin\;\theta}{\cos\;\theta}\) and \(\sec\;\theta = \frac{1}{\cos\;\theta}\), we get:

\[1 ~+~ \tan^2 \;\theta ~=~ \sec^2 \;\theta\]

Likewise, dividing both sides of formula (3) by \(\sin^2 \;\theta\) gives

\[\frac{\cos^2 \;\theta}{\sin^2 \;\theta} ~+~ \frac{\sin^2 \;\theta}{\sin^2 \;\theta} ~=~ \frac{1}{\sin^2 \;\theta} ~~,\]

so since \(\cot\;\theta = \frac{\cos\;\theta}{\sin\;\theta}\) and \(\csc\;\theta = \frac{1}{\sin\;\theta}\), we get:

\[\cot^2 \;\theta ~+~ 1 ~=~ \csc^2 \;\theta\]

Example 3.1

简化 \(\;\cos^2 \;\theta ~ \tan^2 \;\theta\;\)。

解答： 我们可以使用公式 (1) 来简化：

Simplify \(\;\cos^2 \;\theta ~ \tan^2 \;\theta\;\).

Solution: We can use formula (1) to simplify:

\[\begin{split}\begin{align*} \cos^2 \;\theta~\tan^2 \;\theta ~ &= ~ \cos^2 \;\theta ~\cdot~ \frac{\sin^2 \;\theta}{\cos^2 \;\theta}\\ &= ~ \sin^2 \;\theta \end{align*}\end{split}\]

Example 3.2

简化 \(\;5\sin^2 \;\theta ~+~ 4\cos^2 \;\theta\;\).

解答: 我们可以使用公式 (5) 来简化:

Simplify \(\;5\sin^2 \;\theta ~+~ 4\cos^2 \;\theta\;\).

Solution: We can use formula (5) to simplify:

\[\begin{split}\begin{align*} 5\sin^2 \;\theta ~+~ 4\cos^2 \;\theta ~ &= ~ 5\sin^2 \;\theta ~+~ 4\left( 1 ~-~ \sin^2 \;\theta \right)\\[1mm] &= ~ 5\sin^2 \;\theta ~+~ 4 ~-~ 4\sin^2 \;\theta\\[1mm] &= ~ \sin^2 \;\theta ~+~ 4 \end{align*}\end{split}\]

Example 3.3

证明 \(\;\tan \;\theta ~+~ \cot \;\theta ~=~ \sec \;\theta ~ \csc \;\theta\;\)。

解答： 我们将左边展开，并证明它等于右边：

\[\begin{split}\begin{alignat*}{3} \tan \;\theta + \cot \;\theta ~ &= ~ \frac{\sin\;\theta}{\cos\;\theta} ~+~ \frac{\cos\;\theta}{\sin\;\theta} &{} \qquad &\text{(通过公式 3.1 和 3.2 )}\\[2mm] &= ~ \frac{\sin\;\theta}{\cos\;\theta} \;\cdot\; \frac{\sin\;\theta}{\sin\;\theta} ~+~ \frac{\cos\;\theta}{\sin\;\theta} \;\cdot\; \frac{\cos\;\theta}{\cos\;\theta} &{} \qquad &\text{(将两个分数乘以 $1$)}\\[2mm] &= ~ \frac{\sin^2 \;\theta ~+~ \cos^2 \;\theta}{\cos\;\theta ~ \sin\;\theta} &{} \qquad &\text{(得到共同分母后)}\\[2mm] &= ~ \frac{1}{\cos\;\theta ~ \sin\;\theta} &{} \qquad &\text{(通过公式 3.3)}\\[2mm] &= ~ \frac{1}{\cos\;\theta} ~\cdot~ \frac{1}{\sin\;\theta}\\[2mm] &= ~ \sec \;\theta ~ \csc \;\theta \end{alignat*}\end{split}\]

Prove that \(\;\tan \;\theta ~+~ \cot \;\theta ~=~ \sec \;\theta ~ \csc \;\theta\;\).

Solution: We will expand the left side and show that it equals the right side:

\[\begin{split}\begin{alignat*}{3} \tan \;\theta + \cot \;\theta ~ &= ~ \frac{\sin\;\theta}{\cos\;\theta} ~+~ \frac{\cos\;\theta}{\sin\;\theta} &{} \qquad &\text{(by eq 3.1 and eq 3.2 )}\\[2mm] &= ~ \frac{\sin\;\theta}{\cos\;\theta} \;\cdot\; \frac{\sin\;\theta}{\sin\;\theta} ~+~ \frac{\cos\;\theta}{\sin\;\theta} \;\cdot\; \frac{\cos\;\theta}{\cos\;\theta} &{} \qquad &\text{(multiply both fractions by $1$)}\\[2mm] &= ~ \frac{\sin^2 \;\theta ~+~ \cos^2 \;\theta}{\cos\;\theta ~ \sin\;\theta} &{} \qquad &\text{(after getting a common denominator)}\\[2mm] &= ~ \frac{1}{\cos\;\theta ~ \sin\;\theta} &{} \qquad &\text{(by eq 3.3)}\\[2mm] &= ~ \frac{1}{\cos\;\theta} ~\cdot~ \frac{1}{\sin\;\theta}\\[2mm] &= ~ \sec \;\theta ~ \csc \;\theta \end{alignat*}\end{split}\]

在上面的例子中，我们是如何知道应该展开左边而不是右边的呢？一般来说，虽然这种方法并不总是有效，但恒等式中更复杂的一边可能更容易展开。原因是，由于其复杂性，你可以用这个表达式做更多的事情。例如，如果你被要求证明

\[\sec\;\theta ~-~ \sin\;\theta ~ \tan\;\theta ~=~ \cos\;\theta ~，\]

你用这个恒等式的右边能做的事情不多；它只有一个项 (\(\cos\;\theta\))，没有明显的展开方式。

In the above example, how did we know to expand the left side instead of the right side? In general, though this technique does not always work, the more complicated side of the identity is likely to be easier to expand. The reason is that, by its complexity, there will be more things that you can do with that expression. For example, if you were asked to prove that

\[\sec\;\theta ~-~ \sin\;\theta ~ \tan\;\theta ~=~ \cos\;\theta ~,\]

there would not be much that you could do with the right side of that identity; it consists of a single term (\(\cos\;\theta\)) that offers no obvious means of expansion.

Example 3.4

证明 \(\;\dfrac{1 ~+~ \cot^2 \;\theta}{\sec\;\theta} ~=~ \csc\;\theta ~ \cot\;\theta\;\)。

解答： 两边中，左边看起来更复杂，所以我们展开它：

\[\begin{split}\begin{alignat*}{3} \frac{1 ~+~ \cot^2 \;\theta}{\sec\;\theta} ~ &= ~ \frac{\csc^2 \;\theta}{\sec\;\theta} &{} \qquad &\text{(通过公式 3.11)}\\[1.5mm] &= ~ \dfrac{\csc\;\theta ~\cdot~ \dfrac{1}{\sin\;\theta}}{\dfrac{1}{\cos\;\theta}} &{} &{}\\[2mm] &= ~ \csc\;\theta ~\cdot~ \frac{\cos\;\theta}{\sin\;\theta} &{} &{}\\[2mm] &= ~ \csc \;\theta ~ \cot \;\theta &{} \qquad &\text{(通过公式 3.2)} \end{alignat*}\end{split}\]

Prove that \(\;\dfrac{1 ~+~ \cot^2 \;\theta}{\sec\;\theta} ~=~ \csc\;\theta ~ \cot\;\theta\;\).

Solution: Of the two sides, the left side looks more complicated, so we will expand that:

\[\begin{split}\begin{alignat*}{3} \frac{1 ~+~ \cot^2 \;\theta}{\sec\;\theta} ~ &= ~ \frac{\csc^2 \;\theta}{\sec\;\theta} &{} \qquad &\text{(by eq 3.11)}\\[1.5mm] &= ~ \dfrac{\csc\;\theta ~\cdot~ \dfrac{1}{\sin\;\theta}}{\dfrac{1}{\cos\;\theta}} &{} &{}\\[2mm] &= ~ \csc\;\theta ~\cdot~ \frac{\cos\;\theta}{\sin\;\theta} &{} &{}\\[2mm] &= ~ \csc \;\theta ~ \cot \;\theta &{} \qquad &\text{(by eq 3.2)} \end{alignat*}\end{split}\]

Example 3.5

证明 \(\;\dfrac{\tan^2 \;\theta ~+~ 2}{1 ~+~ \tan^2 \;\theta} ~=~ 1 ~+~ \cos^2 \;\theta\;\)。

解答： 展开左边：

\[\begin{split}\begin{alignat*}{3} \frac{\tan^2 \;\theta ~+~ 2}{1 ~+~ \tan^2 \;\theta} ~ &= ~ \frac{\left( \tan^2 \;\theta ~+~ 1 \right) ~+~ 1}{1 ~+~ \tan^2 \;\theta} &{} \qquad &{}\\[2mm] &= ~ \frac{\sec^2 \;\theta ~+~ 1}{\sec^2 \;\theta} &{} \qquad &\text{(通过公式 3.10)}\\[2mm] &= ~ \frac{\sec^2 \;\theta}{\sec^2 \;\theta} ~+~ \frac{1}{\sec^2 \;\theta} &{} &{}\\[2mm] &= ~ 1 ~+~ \cos^2 \;\theta\ \end{alignat*}\end{split}\]

Prove that \(\;\dfrac{\tan^2 \;\theta ~+~ 2}{1 ~+~ \tan^2 \;\theta} ~=~ 1 ~+~ \cos^2 \;\theta\;\).

Solution: Expand the left side:

\[\begin{split}\begin{alignat*}{3} \frac{\tan^2 \;\theta ~+~ 2}{1 ~+~ \tan^2 \;\theta} ~ &= ~ \frac{\left( \tan^2 \;\theta ~+~ 1 \right) ~+~ 1}{1 ~+~ \tan^2 \;\theta} &{} \qquad &{}\\[2mm] &= ~ \frac{\sec^2 \;\theta ~+~ 1}{\sec^2 \;\theta} &{} \qquad &\text{(by eq 3.10)}\\[2mm] &= ~ \frac{\sec^2 \;\theta}{\sec^2 \;\theta} ~+~ \frac{1}{\sec^2 \;\theta} &{} &{}\\[2mm] &= ~ 1 ~+~ \cos^2 \;\theta\ \end{alignat*}\end{split}\]

当试图证明一个恒等式，其中至少有一边是表达式的比率时， 交叉相乘(cross-multiplying) 可能是一种有效的技巧：

\[\frac{a}{b} ~=~ \frac{c}{d} \quad\text{当且仅当}\quad ad ~=~ bc\]

When trying to prove an identity where at least one side is a ratio of expressions, cross-multiplying can be an effective technique:

\[\frac{a}{b} ~=~ \frac{c}{d} \quad\text{if and only if}\quad ad ~=~ bc\]

Example 3.6

证明 \(\;\dfrac{1 ~+~ \sin\;\theta}{\cos\;\theta} ~=~ \dfrac{\cos\;\theta}{1 ~-~ \sin\;\theta}\;\)。

解答： 对等式两边进行交叉相乘并约简，直到它们相等为止：

Prove that \(\;\dfrac{1 ~+~ \sin\;\theta}{\cos\;\theta} ~=~ \dfrac{\cos\;\theta}{1 ~-~ \sin\;\theta}\;\).

Solution: Cross-multiply and reduce both sides until it is clear that they are equal:

\[\begin{split}\begin{align*} ( 1 ~+~ \sin\;\theta ) ( 1 ~-~ \sin\;\theta ) ~ &= ~ \cos\;\theta ~\cdot~ \cos\;\theta\\ 1 ~-~ \sin^2 \;\theta ~ &= ~ \cos^2 \;\theta \end{align*}\end{split}\]

根据公式 (5)，最后一个等式的两边确实相等。因此，原始恒等式成立。

By (5) both sides of the last equation are indeed equal. Thus, the original identity holds.

Example 3.7

假设 \(\;a\,\cos\;\theta = b\;\) 且 \(\;c\,\sin\;\theta = d\;\)，其中 \(\theta\) 为某个角度， 且有常数 $a$、$b$、$c$ 和 $d$。证明 \(\;a^2 c^2 = b^2 c^2 + a^2 d^2\)。

解法： 将第一个等式两边乘以 $c$，将第二个等式乘以 $a$：

\[\begin{split}\begin{align*} ac\,\cos\;\theta ~ &= ~ bc\\ ac\,\sin\;\theta ~ &= ~ ad \end{align*}\end{split}\]

现在对上述每个方程求平方，然后将它们相加得到：

\[\begin{split}\begin{align*} (ac\,\cos\;\theta)^2 ~+~ (ac\,\sin\;\theta)^2 ~ &= ~ (bc)^2 ~+~ (ad)^2\\ (ac)^2 \left( \cos^2 \;\theta ~+~ \sin^2 \;\theta \right)~ &= ~ b^2 c^2 ~+~ a^2 d^2\\ a^2 c^2 ~ &= ~ b^2 c^2 ~+~ a^2 d^2 \qquad\text{(通过公式 3.3)} \end{align*}\end{split}\]

注意： \(\theta\) 并没有出现在我们的最终结果中。诀窍在于求出 \(\cos\;\theta\;\) 和 \(\sin\;\theta\;\) 的公共系数 ($ac$)，这样我们就可以使用 \(\cos^2 \theta + \sin^2 \theta = 1\) 。这是一种从方程组中消除三角函数的常用技巧。

Suppose that \(\;a\,\cos\;\theta = b\;\) and \(\;c\,\sin\;\theta = d\;\) for some angle \(\theta\) and some constants $a$, $b$, $c$, and $d$. Show that \(\;a^2 c^2 = b^2 c^2 + a^2 d^2\).

Solution: Multiply both sides of the first equation by $c$ and the second equation by $a$:

\[\begin{split}\begin{align*} ac\,\cos\;\theta ~ &= ~ bc\\ ac\,\sin\;\theta ~ &= ~ ad \end{align*}\end{split}\]

Now square each of the above equations then add them together to get:

\[\begin{split}\begin{align*} (ac\,\cos\;\theta)^2 ~+~ (ac\,\sin\;\theta)^2 ~ &= ~ (bc)^2 ~+~ (ad)^2\\ (ac)^2 \left( \cos^2 \;\theta ~+~ \sin^2 \;\theta \right)~ &= ~ b^2 c^2 ~+~ a^2 d^2\\ a^2 c^2 ~ &= ~ b^2 c^2 ~+~ a^2 d^2 \qquad\text{(by 3.3)} \end{align*}\end{split}\]

Notice how \(\theta\) does not appear in our final result. The trick was to get a common coefficient ($ac$) for \(\;\cos\;\theta\;\) and \(\;\sin\;\theta\;\) so that we could use \(\;\cos^2 \;\theta + \sin^2 \;\theta = 1\). This is a common technique for eliminating trigonometric functions from systems of equations.

练习

Exercises

1. 我们证明了，对于所有 \(\theta\)，\(\sin\;\theta ~=~ \pm\,\sqrt{1 ~-~ \cos^2 \;\theta}\;\)。请给出一个角 \(\theta\) 的例子，使得 \(\sin\;\theta ~=~ -\sqrt{1 ~-~ \cos^2 \;\theta}\;\)。

2. 我们证明了，对于所有 \(\theta\)，\(\cos\;\theta ~=~ \pm\,\sqrt{1 ~-~ \sin^2 \;\theta}\;\)。举一个角度 \(\theta\) 的例子，使得 \(\cos\;\theta ~=~ -\sqrt{1 ~-~ \sin^2 \;\theta}\;\)。

3. 假设给定一个如下形式的二元方程组： [1]

   \[\begin{split}\begin{align*} A\,\cos\;\phi ~ &= ~ B\, \nu_1 ~-~ B\nu_2 \;\cos\;\theta\\ A\,\sin\;\phi ~ &= ~ B\, \nu_2 \;\sin\;\theta ~。 \end{align*}\end{split}\]

证明 \(\;A ^2 ~=~ B^2 \left( \nu_1^2 ~+~ \nu_2^2 ~-~ 2\nu_1 \nu_2 \;\cos\theta\ \right)\)。

[1] (1,2)

这类方程常见于物理学中，例如光子-电子碰撞研究。参见 W. Rindler 所著《狭义相对论》第 95-97 页，爱丁堡：Oliver and Boyd 出版社，1960 年。

These types of equations arise in physics, e.g. in the study of photon-electron collisions. See pp. 95-97 in W. Rindler, Special Relativity, Edinburgh: Oliver and Boyd, LTD., 1960.

练习 4-16 证明给定的恒等式。

4. \(\cos\;\theta ~ \tan\;\theta ~=~ \sin\;\theta\)

5. \(\sin\;\theta ~ \cot\;\theta ~=~ \cos\;\theta\)

6. \(\dfrac{\tan\;\theta}{\cot\;\theta} ~=~ \tan^2 \;\theta\)

7. \(\dfrac{\csc\;\theta}{\sin\;\theta} ~=~ \csc^2 \;\theta\)

8. \(\dfrac{\cos^2 \;\theta}{1 ~+~ \sin\;\theta} ~=~ 1 ~-~ \sin\;\theta\)

9. \(\dfrac{1 ~-~ 2\;\cos^2 \;\theta}{\sin\;\theta ~ \cos\;\theta} ~=~ \tan\;\theta ~-~ \cot\;\theta\)

10. \(\sin^4 \;\theta ~-~ \cos^4 \;\theta ~=~ \sin^2 \;\theta ~-~ \cos^2 \;\theta\)

11. \(\cos^4 \;\theta ~-~ \sin^4 \;\theta ~=~ 1 ~-~ 2\;\sin^2 \;\theta\)

12. \(\dfrac{1 ~-~ \tan\;\theta}{1 ~+~ \tan\;\theta} ~=~ \dfrac{\cot\;\theta ~-~ 1}{\cot\;\theta ~+~ 1}\)

13. \(\dfrac{\tan\;\theta ~+~ \tan\;\phi}{\cot\;\theta ~+~ \cot\;\phi} ~=~ \tan\;\theta ~ \tan\;\phi\)

14. \(\dfrac{\sin^2 \;\theta}{1 ~-~ \sin^2 \;\theta} ~=~ \tan^2 \;\theta\)

15. \(\dfrac{1 ~-~ \tan^2 \;\theta}{1 ~-~ \cot^2 \;\theta} ~=~ 1 ~-~ \sec^2 \;\theta\)

16. \(\sin\;\theta ~=~ \pm\,\dfrac{\tan\;\theta}{\sqrt{1 ~+~ \tan^2 \;\theta}}\qquad\) (提示：在练习 14 中求解 \(\;\sin^2 \theta\;\)。})

Figure 3.1.2

17. 有时恒等式可以用几何方法证明。例如，为了证明练习 16 中的恒等式，在 QI 中画一个锐角 \(\theta\)，并在其端点取点 $(1,y)$，如图 3.1.2 所示。$y$ 必须等于多少？用这个值来证明锐角 \(\theta\) 的恒等式。解释你需要在图 3.1.2 中做出哪些调整才能在其他象限中证明 \(\theta\) 的恒等式。如果 \(\theta\) 在任一轴上，恒等式是否成立？

18. 与练习 16 类似，仅根据 \(\tan\;\theta\) 找到 \(\cos\;\theta\) 的表达式。

19. 仅根据 \(\sin\;\theta\) 找到 \(\tan\;\theta\) 的表达式，以及仅根据 \(\cos\;\theta\) 找到 \(\tan\;\theta\) 的表达式

20. 假设坐标为 \((x,y)=(a\;(\cos\;\psi\;-\;\epsilon),a\sqrt{1 - \epsilon^2}~\sin\;\psi)\) 的点到原点的距离为 $r>0$，其中 $a>0$ 且 \(0 < \epsilon < 1\)。使用 \(\;r^2 = x^2 + y^2\) 可证明 \(\;r = a\;(1 \;-\; \epsilon\;\cos\;\psi)\)。（注：这些坐标出现在行星椭圆轨道的研究中。）

21. 证明每个三角函数都可以用正弦函数来表示。

1. We showed that \(\;\sin\;\theta ~=~ \pm\,\sqrt{1 ~-~ \cos^2 \;\theta}\;\) for all \(\theta\). Give an example of an angle \(\theta\) such that \(\sin\;\theta ~=~ -\sqrt{1 ~-~ \cos^2 \;\theta}\;\).

2. We showed that \(\;\cos\;\theta ~=~ \pm\,\sqrt{1 ~-~ \sin^2 \;\theta}\;\) for all \(\theta\). Give an example of an angle \(\theta\) such that \(\cos\;\theta ~=~ -\sqrt{1 ~-~ \sin^2 \;\theta}\;\).

3. Suppose that you are given a system of two equations of the following form: [1]

   \[\begin{split}\begin{align*} A\,\cos\;\phi ~ &= ~ B\, \nu_1 ~-~ B\nu_2 \;\cos\;\theta\\ A\,\sin\;\phi ~ &= ~ B\, \nu_2 \;\sin\;\theta ~. \end{align*}\end{split}\]

   Show that \(\;A ^2 ~=~ B^2 \left( \nu_1^2 ~+~ \nu_2^2 ~-~ 2\nu_1 \nu_2 \;\cos\theta\ \right)\).

For Exercises 4-16, prove the given identity.

4. \(\cos\;\theta ~ \tan\;\theta ~=~ \sin\;\theta\)

5. \(\sin\;\theta ~ \cot\;\theta ~=~ \cos\;\theta\)

6. \(\dfrac{\tan\;\theta}{\cot\;\theta} ~=~ \tan^2 \;\theta\)

7. \(\dfrac{\csc\;\theta}{\sin\;\theta} ~=~ \csc^2 \;\theta\)

8. \(\dfrac{\cos^2 \;\theta}{1 ~+~ \sin\;\theta} ~=~ 1 ~-~ \sin\;\theta\)

9. \(\dfrac{1 ~-~ 2\;\cos^2 \;\theta}{\sin\;\theta ~ \cos\;\theta} ~=~ \tan\;\theta ~-~ \cot\;\theta\)

10. \(\sin^4 \;\theta ~-~ \cos^4 \;\theta ~=~ \sin^2 \;\theta ~-~ \cos^2 \;\theta\)

11. \(\cos^4 \;\theta ~-~ \sin^4 \;\theta ~=~ 1 ~-~ 2\;\sin^2 \;\theta\)

12. \(\dfrac{1 ~-~ \tan\;\theta}{1 ~+~ \tan\;\theta} ~=~ \dfrac{\cot\;\theta ~-~ 1}{\cot\;\theta ~+~ 1}\)

13. \(\dfrac{\tan\;\theta ~+~ \tan\;\phi}{\cot\;\theta ~+~ \cot\;\phi} ~=~ \tan\;\theta ~ \tan\;\phi\)

14. \(\dfrac{\sin^2 \;\theta}{1 ~-~ \sin^2 \;\theta} ~=~ \tan^2 \;\theta\)

15. \(\dfrac{1 ~-~ \tan^2 \;\theta}{1 ~-~ \cot^2 \;\theta} ~=~ 1 ~-~ \sec^2 \;\theta\)

16. \(\sin\;\theta ~=~ \pm\,\dfrac{\tan\;\theta}{\sqrt{1 ~+~ \tan^2 \;\theta}}\qquad\) (Hint: Solve for \(\;\sin^2 \theta\;\) in Exercise 14.})

Figure 3.1.2

17. Sometimes identities can be proved by geometrical methods. For example, to prove the identity in Exercise 16, draw an acute angle \(\theta\) in QI and pick the point $(1,y)$ on its terminal side, as in Figure 3.1.2. What must $y$ equal? Use that to prove the identity for acute \(\theta\). Explain the adjustment(s) you would need to make in Figure 3.1.2 to prove the identity for \(\theta\) in the other quadrants. Does the identity hold if \(\theta\) is on either axis?

18. Similar to Exercise 16 , find an expression for \(\cos\;\theta\) solely in terms of \(\tan\;\theta\).

19. Find an expression for \(\tan\;\theta\) solely in terms of \(\sin\;\theta\), and one solely in terms of \(\cos\;\theta\)

20. Suppose that a point with coordinates \((x,y)=(a\;(\cos\;\psi\;-\;\epsilon),a\sqrt{1 - \epsilon^2}~\sin\;\psi)\) is a distance $r>0$ from the origin, where $a>0$ and \(0 < \epsilon < 1\). Use \(\;r^2 = x^2 + y^2\) to show that \(\;r = a\;(1 \;-\; \epsilon\;\cos\;\psi)\;\). (Note: These coordinates arise in the study of elliptical orbits of planets.)

21. Show that each trigonometric function can be put in terms of the sine function.

2. 和差公式

Sum and Difference Formulas

We will now derive identities for the trigonometric functions of the sum and difference of two angles. For the sum of any two angles $A$ and $B$, we have the addition formulas:

(12)\[\sin\;(A+B) ~=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B\]

(13)\[\cos\;(A+B) ~=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B\]

To prove these, first assume that $A$ and $B$ are acute angles. Then $A+B$ is either acute or obtuse, as in Figure 3.2.1. Note in both cases that \(\angle\,QPR = A\), since

\[\begin{split}\begin{align*} \angle\,QPR ~&=~ \angle\,QPO - \angle\,OPM ~=~ (90^\circ - B) - (90^\circ - (A+B)) ~=~ A ~~\text{in Figure 3.2.1 (a), and}\\ \angle\,QPR ~&=~ \angle\,QPO + \angle\,OPM ~=~ (90^\circ - B) + (90^\circ - (180^\circ - (A+B))) ~=~ A ~~\text{in Figure 3.2.1 (b).} \end{align*}\end{split}\]

Figure 3.2.1 \(\sin\;(A+B)\) and \(\cos\;(A+B)\) for acute $A$ and $B$

Thus,

(14)\[\begin{split}\begin{align} \sin\;(A+B) ~&=~ \frac{MP}{OP} ~=~ \frac{MR+RP}{OP} ~=~ \frac{NQ+RP}{OP} ~=~ \frac{NQ}{OP} ~+~ \frac{RP}{OP}\notag\\ &=~ \frac{NQ}{OQ}\,\cdot\,\frac{OQ}{OP} ~+~ \frac{RP}{PQ}\,\cdot\,\frac{PQ}{OP}\notag\\ &=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B ~, \end{align}\end{split}\]

and

(15)\[\begin{split}\begin{align} \cos\;(A+B) ~&=~ \frac{OM}{OP} ~=~ \frac{ON-MN}{OP} ~=~ \frac{ON-RQ}{OP} ~=~ \frac{ON}{OP} ~-~ \frac{RQ}{OP}\notag\\ &=~ \frac{ON}{OQ}\,\cdot\,\frac{OQ}{OP} ~-~ \frac{RQ}{PQ}\,\cdot\,\frac{PQ}{OP}\notag\\ &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~. \end{align}\end{split}\]

So we have proved the identities for acute angles $A$ and $B$. It is simple to verify that they hold in the special case of \(A=B=0^\circ\). For general angles, we will need to use the relations we derived in Section 1.5 which involve adding or subtracting \(90^\circ\):

\[\begin{split}\begin{alignat*}{4} \sin\;(\theta + 90^\circ) ~ &= ~ \phantom{-}\cos\;\theta &\qquad\quad \sin\;(\theta - 90^\circ) ~ &= ~ -\cos\;\theta\\ \cos\;(\theta + 90^\circ) ~ &= ~ -\sin\;\theta &\qquad\quad \cos\;(\theta - 90^\circ) ~ &= ~ \phantom{-}\sin\;\theta \end{alignat*}\end{split}\]

These will be useful because any angle can be written as the sum of an acute angle (or \(0^\circ\)) and integer multiples of \(\pm90^\circ\). For example, \(155^\circ = 65^\circ + 90^\circ\), \(222^\circ = 42^\circ + 2(90^\circ)\), \(-77^\circ = 13^\circ - 90^\circ\), etc. So if we can prove that the identities hold when adding or subtracting \(90^\circ\) to or from either $A$ or $B$, respectively, where $A$ and $B$ are acute or \(0^\circ\), then the identities will also hold when repeatedly adding or subtracting \(90^\circ\), and hence will hold for all angles. Replacing $A$ by \(A+90^\circ\) and using the relations for adding \(90^\circ\) gives

\[\begin{split}\begin{align*} \sin\;((A+90^\circ) + B) ~&=~ \sin\;((A+B) + 90^\circ) ~=~ \cos\;(A+B)~,\\ &=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B ~~\text{(by equation 3.15)}\\ &=~ \sin\;(A + 90^\circ)~\cos\;B ~+~ \cos\;(A + 90^\circ)~\sin\;B ~, \end{align*}\end{split}\]

so the identity holds for \(A+90^\circ\) and $B$ (and, similarly, for $A$ and \(B+90^\circ\)). Likewise,

\[\begin{split}\begin{align*} \sin\;((A-90^\circ) + B) ~&=~ \sin\;((A+B) - 90^\circ) ~=~ -\cos\;(A+B)~,\\ &=~ -(\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B) \\ &=~ (-\cos\;A) ~ \cos\;B ~+~ \sin\;A ~ \sin\;B\\ &=~ \sin\;(A - 90^\circ)~\cos\;B ~+~ \cos\;(A - 90^\circ)~\sin\;B ~, \end{align*}\end{split}\]

so the identity holds for \(A-90^\circ\) and $B$ (and, similarly, for $A$ and \(B+90^\circ\)). Thus, the addition formula (ref{eqn:sumsin}) for sine holds for emph{all} $A$ and $B$. A similar argument shows that the addition formula (ref{eqn:sumcos}) for cosine is true for all $A$ and $B$. [qed]

Replacing $B$ by $-B$ in the addition formulas and using the relations \(\sin\;(-\theta) = -\sin\;\theta\) and \(\cos\;(-\theta) = \cos\;\theta\) from Section 1.5 gives us the subtraction formulas :

(16)\[\sin\;(A-B) ~=~ \sin\;A ~ \cos\;B ~-~ \cos\;A ~ \sin\;B\label{}\]

(17)\[\cos\;(A-B) ~=~ \cos\;A ~ \cos\;B ~+~ \sin\;A ~ \sin\;B\label{eqn:}\]

Using the identity \(\tan\;\theta = \frac{\sin\;\theta}{\cos\;\theta}\), and the addition formulas for sine and cosine, we can derive the addition formula for tangent:

\[\begin{split}\begin{align*} \tan\;(A+B) ~&=~ \frac{\sin\;(A+B)}{\cos\;(A+B)}\\[5pt] &=~ \frac{\sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B}{\cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B}\\[5pt] &=~ \frac{\dfrac{\sin\;A ~ \cos\;B}{\cos\;A ~ \cos\;B} ~+~ \dfrac{\cos\;A ~ \sin\;B}{\cos\;A ~ \cos\;B}}{\dfrac{\cos\;A ~ \cos\;B}{\cos\;A ~ \cos\;B} ~-~ \dfrac{\sin\;A ~ \sin\;B}{\cos\;A ~ \cos\;B}}\quad\text{(divide top and bottom by $\cos\;A ~ \cos\;B$)}\\[5pt] &=~ \frac{\dfrac{\sin\;A}{\cos\;A} \;\cdot\; \cancel{\dfrac{\cos\;B}{\cos\;B}} ~+~ \cancel{\dfrac{\cos\;A}{\cos\;A}} \;\cdot\; \dfrac{\sin\;B}{\cos\;B}}{1 ~-~ \dfrac{\sin\;A}{\cos\;A} \;\cdot\; \dfrac{\sin\;B}{\cos\;B}} ~=~ \frac{\tan\;A ~+~ \tan\;B}{1 ~-~ \tan\;A ~ \tan\;B} \end{align*}\end{split}\]

This, combined with replacing $B$ by $-B$ and using the relation \(\tan\;(-\theta) = -\tan\;\theta\), gives us the addition and subtraction formulas for tangent:

(18)\[\tan\;(A+B) ~=~ \frac{\tan\;A ~+~ \tan\;B}{1 ~-~ \tan\;A ~ \tan\;B}\]

(19)\[\tan\;(A-B) ~=~ \frac{\tan\;A ~-~ \tan\;B}{1 ~+~ \tan\;A ~ \tan\;B}\]

Example 3.8

Given angles $A$ and $B$ such that \(\sin\;A = \frac{4}{5}\), \(\cos\;A = \frac{3}{5}\), \(\sin\;B = \frac{12}{13}\), and \(\cos\;B = \frac{5}{13}\), find the exact values of \(\sin\;(A+B)\), \(\cos\;(A+B)\), and \(\tan\;(A+B)\).

Solution: Using the addition formula for sine, we get:

\[\begin{split}\begin{align*} \sin\;(A+B) ~&=~ \sin\;A ~ \cos\;B ~+~ \cos\;A ~ \sin\;B\\ &=~ \frac{4}{5} \;\cdot\; \frac{5}{13} ~+~ \frac{3}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\sin\;(A+B) ~=~ \frac{56}{65}}\\ \end{align*}\end{split}\]

Using the addition formula for cosine, we get

\[\begin{split}\begin{align*} \cos\;(A+B) ~&=~ \cos\;A ~ \cos\;B ~-~ \sin\;A ~ \sin\;B\\ &=~ \frac{3}{5} \;\cdot\; \frac{5}{13} ~-~ \frac{4}{5} \;\cdot\; \frac{12}{13} \quad\Rightarrow\quad \boxed{\cos\;(A+B) ~=~ -\frac{33}{65}}\\ \end{align*}\end{split}\]

Instead of using the addition formula for tangent, we can use the results above:

\[\begin{align*} \tan\;(A+B) ~&=~ \frac{\sin\;(A+B)}{\cos\;(A+B)} ~=~ \frac{\frac{56}{65}}{-\frac{33}{65}} \quad\Rightarrow\quad \boxed{\tan\;(A+B) ~=~ -\frac{56}{33}} \end{align*}\]

Example 3.9

Prove the following identity:

\[\sin\;(A+B+C) ~=~ \sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C \;+\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C\]

Solution: Treat $A+B+C$ as $(A+B)+C$ and use the addition formulas three times:

\[\begin{split}\begin{align*} \sin\;(A+B+C) ~&=~ \sin\;((A+B)+C)\\ &=~ \sin\;(A+B)~\cos\;C \;+\; \cos\;(A+B)~\sin\;C\\ &=~ (\sin\;A ~ \cos\;B \;+\; \cos\;A ~ \sin\;B)~\cos\;C \;+\; (\cos\;A ~ \cos\;B \;-\; \sin\;A ~ \sin\;B)~\sin\;C\\ &=~ \sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C \;+\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C \end{align*}\end{split}\]

Example 3.10

For any triangle \(\triangle\,ABC\), show that \(\tan\;A + \tan\;B + \tan\;C = \tan\;A~\tan\;B~\tan\;C\).

Solution: Note that this is not an identity which holds for all angles; since $A$, $B$, and $C$ are the angles of a triangle, it holds when $A$, $B$, $C$ \(> 0^\circ\) and \(A + B + C = 180^\circ\). So using \(C = 180^\circ - (A+B)\) and the relation \(\;\tan\;(180^\circ - \theta) = -\tan\;\theta\;\) from Section 1.5, we get:

\[\begin{split}\begin{align*} \tan\;A \;+\; \tan\;B \;+\; \tan\;C ~&=~ \tan\;A \;+\; \tan\;B \;+\; \tan\;(180^\circ - (A+B))\\ &=~ \tan\;A \;+\; \tan\;B \;-\; \tan\;(A+B)\\ &=~ \tan\;A \;+\; \tan\;B \;-\; \frac{\tan\;A + \tan\;B}{1 - \tan\;A ~ \tan\;B}\\ &=~ (\tan\;A \;+\; \tan\;B)~\left( 1 \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right)\\ &=~ (\tan\;A \;+\; \tan\;B)~\left( \dfrac{1 - \tan\;A ~ \tan\;B}{1 - \tan\;A ~ \tan\;B} \;-\; \dfrac{1}{1 - \tan\;A ~ \tan\;B} \right)\\ &=~ (\tan\;A \;+\; \tan\;B)\;\cdot\;\left( \frac{-\tan\;A ~ \tan\;B}{{1 - \tan\;A ~ \tan\;B}} \right)\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; \left( -\frac{\tan\;A \;+\; \tan\;B}{{1 - \tan\;A ~ \tan\;B}} \right)\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; (-\tan\;(A+B))\\ &=~ \tan\;A ~ \tan\;B \;\cdot\; (\tan\;(180^\circ - (A+B)))\\ &=~ \tan\;A ~ \tan\;B ~ \tan\;C \end{align*}\end{split}\]

Example 3.11

Let $A$, $B$, $C$, and $D$ be positive angles such that \(A+B+C+D=180^\circ\). Show that [2]

\[\sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~=~ \sin\;(A+C)~\sin\;(B+C) ~.\]

[2]

This is the “trigonometric form” of Ptolemy’s Theorem, which says that a quadrilateral can be inscribed in a circle if and only if the sum of the products of its opposite sides equals the product of its diagonals.

Solution: It may be tempting to expand the right side, since it appears more complicated. However, notice that the right side has no $D$ term. So instead, we will expand the left side, since we can eliminate the $D$ term on that side by using \(D=180^\circ - (A+B+C)\) and the relation

\[\sin\;(180^\circ -(A+B+C)) ~=~ \sin\;(A+B+C).\]

So since \(\;\sin\;D = \sin\;(A+B+C)\), we get

\[\begin{split}\begin{align*} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~&=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;(A+B+C) ~,~\text{so by Example 3.9 we get}\\ &=~ \sin\;A~\sin\;B ~+~ \sin\;C~(\sin\;A~\cos\;B~\cos\;C \;+\; \cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +\; \cos\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\sin\;C)\\ &=~ \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;A~\cos\;B~\cos\;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \sin\;C~\cos\;A~\cos\;B~\sin\;C ~-~ \sin\;C~\sin\;A~\sin\;B~\sin\;C ~.\\ \end{align*}\end{split}\]

It may not be immediately obvious where to go from here, but it is not completely guesswork. We need to end up with \(\sin\;(A+C)~\sin\;(B+C)\), and we know that \(\sin\;(B+C) = \sin\;B~\cos\;C + \cos\;B~\sin\;C\). There are two terms involving \(\;\cos\;B~\sin\;C\), so group them together to get

\[\begin{split}\begin{align*} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~ &=~ \sin\;A~\sin\;B ~-~ \sin\;C~\sin\;A~\sin\;B~\sin\;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~(\sin\;A~\cos\;C ~+~ \cos\;A~\sin\;C)\\ &=~ \sin\;A~\sin\;B~(1 - \sin^2 \;C) ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)\\ &=~ \sin\;A~\sin\;B~\cos^2 \;C ~+~ \sin\;C~\cos\;A~\sin\;B~\cos\;C\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)~.\\ \end{align*}\end{split}\]

We now have two terms involving \(\;\sin\;B~\cos\;C\), which we can factor out:

\[\begin{split}\begin{align*} \sin\;A~\sin\;B ~+~ \sin\;C~\sin\;D ~ &=~ \sin\;B~\cos\;C~(\sin\;A~\cos\;C + \cos\;A~\sin\;C~)\\ &\mathrel{\phantom{=}} {} +~ \cos\;B~\sin\;C~\sin\;(A+C)\\ &=~ \sin\;B~\cos\;C~\sin\;(A+C) ~+~ \cos\;B~\sin\;C~\sin\;(A+C)\\ &=~ \sin\;(A+C)~(\sin\;B~\cos\;C + \cos\;B~\sin\;C)\\ &=~ \sin\;(A+C)~\sin\;(B+C) \end{align*}\end{split}\]

Example 3.12

In the study of the propagation of electromagnetic waves, Snell’s law gives the relation

(20)\[n_1 ~\sin\;\theta_1 ~=~ n_2 ~\sin\;\theta_2 ~,\]

where \(\theta_1\) is the angle of incidence at which a wave strikes the planar boundary between two mediums, \(\theta_2\) is the angle of transmission of the wave through the new medium, and $n_1$ and $n_2$ are the indexes of refraction of the two mediums. The quantity

(21)\[r_{1\;2\;s} ~=~ \frac{n_1 ~\cos\;\theta_1 ~-~ n_2 ~\cos\;\theta_2}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2}\]

is called the Fresnel coefficient for normal incidence reflection of the wave for s-polarization. Show that this can be written as:

\[r_{1\;2\;s} ~=~ \frac{\sin\;(\theta_2 - \theta_1)}{\sin\;(\theta_2 + \theta_1)}\]

Solution: Multiply the top and bottom of \(r_{1\;2\;s}\) by \(\;\sin\;\theta_1 ~ \sin\;\theta_2\;\) to get:

\[\begin{split}\begin{align*} r_{1\;2\;s} ~&=~ \frac{n_1 ~\cos\;\theta_1 ~-~ n_2 ~\cos\;\theta_2}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2} \;\cdot\; \frac{\sin\;\theta_1 ~ \sin\;\theta_2}{\sin\;\theta_1 ~ \sin\;\theta_2}\\[7pt] &=~ \frac{(n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~-~ (n_2 ~\sin\;\theta_2)~\cos\;\theta_2 ~\sin\;\theta_1}{ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~+~ (n_2 ~\sin\;\theta_2)~\cos\;\theta_2 ~\sin\;\theta_1}\\[7pt] &=~ \frac{(n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~-~ (n_1 ~\sin\;\theta_1)~\cos\;\theta_2 ~\sin\;\theta_1}{ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_1 ~+~ (n_1 ~\sin\;\theta_1)~\cos\;\theta_2 ~\sin\;\theta_1} \qquad\text{(by Snell's law)}\\[7pt] &=~ \frac{\sin\;\theta_2 ~\cos\;\theta_1 ~-~ \cos\;\theta_2 ~\sin\;\theta_1}{ \sin\;\theta_2 ~\cos\;\theta_1 ~+~ \cos\;\theta_2 ~\sin\;\theta_1}\\[7pt] &=~ \frac{\sin\;(\theta_2 - \theta_1)}{\sin\;(\theta_2 + \theta_1)} \end{align*}\end{split}\]

The last two examples demonstrate an important aspect of how identities are used in practice: recognizing terms which are part of known identities, so that they can be factored out. This is a common technique.

练习

Exercises

1. Verify the addition formulas (12) and (13) for \(A=B=0^\circ\).

For Exercises 2 and 3, find the exact values of \(\sin\;(A+B)\), \(\cos\;(A+B)\), and \(\tan\;(A+B)\).

2. \(\sin\;A = \frac{8}{17}, \cos\;A = \frac{15}{17}, \sin\;B = \frac{24}{25}, \cos\;B = \frac{7}{25}\)

3. \(\sin\;A = \frac{40}{41}, \cos\;A = \frac{9}{41}, \sin\;B = \frac{20}{29}, \cos\;B = \frac{21}{29}\)

4. Use \(75^\circ = 45^\circ + 30^\circ\) to find the exact value of \(\;\sin\;75^\circ\).

5. Use \(15^\circ = 45^\circ - 30^\circ\) to find the exact value of \(\;\tan\;15^\circ\).

6. Prove the identity $;sin;theta + cos;theta = sqrt{2};sin;(theta + 45^circ);$. Explain why this shows that

   \[-\sqrt{2} ~\le~ \;\sin\;\theta ~+~ \cos\;\theta ~\le~ \sqrt{2}\]

   for all angles \(\theta\). For which \(\theta\) between \(0^\circ\) and \(360^\circ\) would \(\;\sin\;\theta \;+\; \cos\;\theta\;\) be the largest?

For Exercises ref{exer:iden32start}-ref{exer:iden32end}, prove the given identity.

7. \(\cos\;(A+B+C) \;=\; \cos\;A~\cos\;B~\cos\;C \;-\; \cos\;A~\sin\;B~\sin\;C \;-\; \sin\;A~\cos\;B~\sin\;C \;-\; \sin\;A~\sin\;B~\cos\;C\)

8. \(\tan\;(A+B+C) ~=~ \dfrac{\tan\;A \;+\; \tan\;B \;+\; \tan\;C \;-\; \tan\;A~\tan\;B~\tan\;C}{1 \;-\; \tan\;B~\tan\;C \;-\; \tan\;A~\tan\;C \;-\; \tan\;A~\tan\;B}\)

9. \(\cot\;(A+B) ~=~ \dfrac{\cot\;A~\cot\;B \;-\; 1}{\cot\;A \;+\; \cot\;B}\)

10. \(\cot\;(A-B) ~=~ \dfrac{\cot\;A~\cot\;B \;+\; 1}{\cot\;B \;-\; \cot\;A}\)

11. \(\tan\;(\theta + 45^\circ) ~=~ \dfrac{1 \;+\; \tan\;\theta}{1 \;-\; \tan\;\theta}\)

12. \(\dfrac{\cos\;(A+B)}{\sin\;A~\cos\;B} ~=~ \cot\;A \;-\; \tan\;B\)

13. \(\cot\;A ~+~ \cot\;B ~=~ \dfrac{\sin\;(A+B)}{\sin\;A~\sin\;B}\)

14. \(\dfrac{\sin\;(A-B)}{\sin\;(A+B)} ~=~ \dfrac{\cot\;B \;-\; \cot\;A}{\cot\;B \;+\; \cot\;A}\)

15. Generalize Exercise 6: For any $a$ and $b$, \(-\sqrt{a^2 + b^2} \;\le\; a\;\sin\;\theta \;+\; b\;\cos\;\theta \;\le\; \sqrt{a^2 + b^2}\;\) for all \(\theta\).

16. Continuing Example 3.12, use Snell’s law to show that the s-polarization transmission Fresnel coefficient

    (22)\[t_{1\;2\;s} ~=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_1 ~\cos\;\theta_1 ~+~ n_2 ~\cos\;\theta_2}\]

    can be written as:

    \[t_{1\;2\;s} ~=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_2 + \theta_1)}\]

17. Suppose that two lines with slopes $m_1$ and $m_2$, respectively, intersect at an angle \(\theta\) and are not perpendicular (i.e. \(\theta \ne 90^\circ\)), as in the figure on the right. Show that

    \[\tan\;\theta ~=~ \left| \frac{m_1 ~-~ m_2}{1 ~+~ m_1 \; m_2} \right| ~.\]

    (Hint: Use Example 3.12 from Section 1.5. )

18. Use Exercise 3.12 to find the angle between the lines $y=2x+3$ and $y=-5x-4$.

19. For any triangle \(\triangle\,ABC\), show that \(\;\cot\;A~\cot\;B ~+~ \cot\;B~\cot\;C ~+~ \cot\;C~\cot\;A ~=~ 1\). (Hint: Use Exercise 9 and \(C=180^\circ - (A+B)\).)

20. For any positive angles $A$, $B$, and $C$ such that \(A+B+C=90^\circ\), show that

\[\tan\;A~\tan\;B ~+~ \tan\;B~\tan\;C ~+~ \tan\;C~\tan\;A ~=~ 1 ~.\]

21. Prove the identity \(\;\sin\;(A+B)~\cos\;B ~-~ \cos\;(A+B)~\sin\;B ~=~ \sin\;A\). Note that the right side depends only on $A$, while the left side depends on both $A$ and $B$.

22. A line segment of length $r > 0$ from the origin to the point $(x,y)$ makes an angle \(\alpha\) with the positive $x$-axis, so that \((x,y) = (r\;\cos\;\alpha,r\;\sin\;\alpha)\), as in the figure below. What are the endpoint’s new coordinates $(x’,y’)$ after a counterclockwise rotation by an angle \(\beta\;\)? Your answer should be in terms of $r$, \(\alpha\), and \(\beta\).

3. 倍角与半角公式

Double-Angle and Half-Angle Formulas

A special case of the addition formulas is when the two angles being added are equal, resulting in the double-angle formulas:

(23)\[\sin\;2\theta ~=~ 2\;\sin\;\theta ~ \cos\;\theta\]

(24)\[\cos\;2\theta ~=~ \cos^2 \;\theta ~-~ \sin^2 \;\theta\]

(25)\[\tan\;2\theta ~=~ \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta}\]

To derive the sine double-angle formula, we see that

\[\begin{split}\sin\;2\theta ~=~ \sin\;(\theta+\theta) ~=~ \sin\;\theta ~ \cos\;\theta ~+~ \cos\;\theta ~ \sin\;\theta ~=~ 2\;\sin\;\theta ~ \cos\;\theta~.\\\end{split}\]

Likewise, for the cosine double-angle formula, we have

\[\begin{split}\cos\;2\theta ~=~ \cos\;(\theta+\theta) ~=~ \cos\;\theta~\cos\;\theta ~-~ \sin\;\theta~\sin\;\theta ~=~ \cos^2 \;\theta ~-~ \sin^2 \;\theta~,\\\end{split}\]

and for the tangent we get

\[\tan\;2\theta ~=~ \tan\;(\theta+\theta) ~=~ \frac{\tan\;\theta ~+~ \tan\;\theta}{1 ~-~ \tan\;\theta ~ \tan\;\theta} ~=~ \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta}\]

Using the identities \(\;\sin^2 \;\theta = 1 - \cos^2 \;\theta\) and \(\;\cos^2 \;\theta = 1 - \sin^2 \;\theta\), we get the following useful alternate forms for the cosine double-angle formula:

(26)\[\cos\;2\theta ~=~ 2\;\cos^2 \;\theta ~-~ 1\]

(27)\[\qquad\quad=~ 1 ~-~ 2\;\sin^2 \;\theta\]

Example 3.13

Prove that \(\;\sin\;3\theta ~=~ 3\;\sin\;\theta ~-~ 4\;\sin^3 \;\theta\;\).

Solution: Using \(3\theta = 2\theta + \theta\), the addition formula for sine, and the double-angle formulas (23) and (27), we get:

\[\begin{split}\begin{align*} \sin\;3\theta ~&=~ \sin\;(2\theta+\theta)\\ &=~ \sin\;2\theta~\cos\;\theta ~+~ \cos\;2\theta~\sin\;\theta\\ &=~ (2\;\sin\;\theta~\cos\;\theta)\;\cos\;\theta ~+~ (1 - 2\;\sin^2 \;\theta)\;\sin\;\theta\\ &=~ 2\;\sin\;\theta~\cos^2 \;\theta ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ &=~ 2\;\sin\;\theta\;(1 - \sin^2 \;\theta) ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ &=~ 3\;\sin\;\theta ~-~ 4\;\sin^3 \;\theta \end{align*}\end{split}\]

Example 3.14

Prove that \(\;\sin\;4z ~=~ \dfrac{4\;\tan\;z~(1 - \tan^2 \;z)}{(1 + \tan^2 \;z)^2}\;\).

Solution: Expand the right side and use \(1 + \tan^2 \;z= \sec^2 \;z\,\):

\[\begin{split}\begin{align*} \dfrac{4\;\tan\;z~(1 - \tan^2 \;z)}{(1 + \tan^2 \;z)^2} ~&=~ \dfrac{4 \;\cdot\; \dfrac{\sin\;z}{\cos\;z} \;\cdot\; \left( \dfrac{\cos^2 \;z}{\cos^2 \;z} - \dfrac{\sin^2 \;z}{\cos^2 \;z} \right)}{( \sec^2 \;z )^2}\\[6pt] &=~ \dfrac{4 \;\cdot\; \dfrac{\sin\;z}{\cos\;z} \;\cdot\; \dfrac{\cos\;2z}{\cos^2 \;z}}{\left( \dfrac{1}{\cos^2 \;z} \right)^2}\quad\qquad\text{(by formula (24))}\\[5pt] &=~ (4\;\sin\;z~\cos\;2z)\;\cos\;z\\ &=~ 2\;(2\;\sin\;z~\cos\;z)\;\cos\;2z\\ &=~ 2\;\sin\;2z~\cos\;2z\quad\qquad\text{(by formula (23))}\\ &=~ \sin\;4z\quad\qquad\text{(by formula (23) with } \theta \text{ replaced by } 2z \text{)} \end{align*}\end{split}\]

Note: Perhaps surprisingly, this seemingly obscure identity has found a use in physics, in the derivation of a solution of the sine-Gordon equation in the theory of nonlinear waves. [3]

[3]

See p.331 in L.A. Ostrovsky and A.I.Potapov, Modulated Waves: Theory and Applications, Baltimore: The Johns Hopkins University Press, 1999.

Closely related to the double-angle formulas are the half-angle formulas:

(28)\[\sin^2 \;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{2}\]

(29)\[\cos^2 \;\tfrac{1}{2}\theta ~=~ \frac{1 \;+\; \cos\;\theta}{2}\]

(30)\[\tan^2 \;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}\]

These formulas are just the double-angle formulas rewritten with \(\theta\) replaced by \(\tfrac{1}{2}\theta\):

\[\begin{split}\begin{align*} \cos\;2\theta \;&=\; 1 \;-\; 2\;\sin^2 \;\theta ~\Rightarrow~ \sin^2 \;\theta \;=\; \frac{1 \;-\; \cos\;2\theta}{2} ~\Rightarrow~ \sin^2 \;\tfrac{1}{2}\theta \;=\; \frac{1 \;-\; \cos\;2\,(\tfrac{1}{2}\theta)}{2} \;=\; \frac{1 \;-\; \cos\;\theta}{2}\\ \cos\;2\theta \;&=\; 2\;\cos^2 \;\theta\;-\; 1 ~\Rightarrow~ \cos^2 \;\theta \;=\; \frac{1 \;+\; \cos\;2\theta}{2} ~\Rightarrow~ \cos^2 \;\tfrac{1}{2}\theta \;=\; \frac{1 \;+\; \cos\;2\,(\tfrac{1}{2}\theta)}{2} \;=\; \frac{1 \;+\; \cos\;\theta}{2} \end{align*}\end{split}\]

The tangent half-angle formula then follows easily:

\[\tan^2 \;\tfrac{1}{2}\theta \;=\; \left( \dfrac{\sin\;\tfrac{1}{2}\theta}{\cos\;\tfrac{1}{2}\theta} \right)^2 \;=\; \dfrac{\sin^2 \;\tfrac{1}{2}\theta}{\cos^2 \;\tfrac{1}{2}\theta} \;=\; \dfrac{\tfrac{1 \;-\; \cos\;\theta}{2}}{\tfrac{1 \;+\; \cos\;\theta}{2}} \;=\; \frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}\]

The half-angle formulas are often used (e.g. in calculus) to replace a squared trigonometric function by a nonsquared function, especially when \(2\theta\) is used instead of \(\theta\).

By taking square roots, we can write the above formulas in an alternate form:

(31)\[\sin\;\tfrac{1}{2}\theta ~=~ \pm\;\sqrt{\frac{1 \;-\; \cos\;\theta}{2}}\]

(32)\[\cos\;\tfrac{1}{2}\theta ~=~ \pm\;\sqrt{\frac{1 \;+\; \cos\;\theta}{2}}\]

(33)\[\tan\;\tfrac{1}{2}\theta ~=~ \pm\;\sqrt{\frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}}\]

In the above form, the sign in front of the square root is determined by the quadrant in which the angle \(\tfrac{1}{2}\theta\) is located. For example, if \(\theta=300^\circ\) then \(\tfrac{1}{2}\theta = 150^\circ\) is in QII. So in this case \(\cos\;\tfrac{1}{2}\theta < 0\) and hence we would have \(\cos\;\tfrac{1}{2}\theta = -\;\sqrt{\frac{1 \;+\; \cos\;\theta}{2}}\).

In formula (33), multiplying the numerator and denominator inside the square root by \((1 - \cos\;\theta)\) gives

\[\tan\;\tfrac{1}{2}\theta ~=~ \pm\;\sqrt{\frac{1 - \cos\;\theta}{1 + \cos\;\theta} \,\cdot\, \frac{1 - \cos\;\theta}{1 - \cos\;\theta}} ~=~ \pm\;\sqrt{\frac{(1 - \cos\;\theta)^2}{1 - \cos^2 \;\theta}} ~=~ \pm\;\sqrt{\frac{(1 - \cos\;\theta)^2}{\sin^2 \;\theta}} ~=~ \pm\;\frac{1 - \cos\;\theta}{\sin\;\theta} ~.\]

But \(1 - \cos\;\theta \ge 0\), and it turns out (see Exercise 10) that \(\tan\;\tfrac{1}{2}\theta\) and \(\sin\;\theta\) always have the same sign. Thus, the minus sign in front of the last expression is not possible (since that would switch the signs of \(\tan\;\tfrac{1}{2}\theta\) and \(\sin\;\theta\)), so we have:

(34)\[\tan\;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{\sin\;\theta}\]

Multiplying the numerator and denominator in formula (34) by \(1 + \cos\;\theta\) gives

\[\tan\;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{\sin\;\theta} \;\cdot\; \frac{1 \;+\; \cos\;\theta}{1 \;+\; \cos\;\theta} ~=~ \frac{1 \;-\; \cos^2 \;\theta}{\sin\;\theta\;(1 \;+\; \cos\;\theta)} ~=~ \frac{\sin^2 \;\theta}{\sin\;\theta\;(1 \;+\; \cos\;\theta)} ~,\]

so we also get:

(35)\[\tan\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;+\; \cos\;\theta}\]

Taking reciprocals in formulas (34) and (35) gives:

(36)\[\cot\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;-\; \cos\;\theta} ~=~ \frac{1 \;+\; \cos\;\theta}{\sin\;\theta}\]

Example 3.15

Prove the identity \(\;\sec^2 \;\tfrac{1}{2}\theta ~=~\dfrac{2\;\sec\;\theta}{\sec\;\theta \;+\; 1}\;\).

Solution: Since secant is the reciprocal of cosine, taking the reciprocal of formula (29) for \(\;\cos^2 \;\tfrac{1}{2}\theta\) gives us

\[\sec^2 \;\tfrac{1}{2}\theta ~=~ \frac{2}{1 \;+\; \cos\;\theta} ~=~ \frac{2}{1 \;+\; \cos\;\theta} \;\cdot\; \frac{\sec\;\theta}{\sec\;\theta} ~=~ \frac{2\;\sec\;\theta}{\sec\;\theta \;+\; 1} ~.\]

练习

Exercises

For Exercises 1-8, prove the given identity.

1. \(\cos\;3\theta ~=~ 4\;\cos^3 \;\theta ~-~ 3\;\cos\;\theta\)

2. \(\tan\;\tfrac{1}{2}\theta ~=~ \csc\;\theta ~-~ \cot\;\theta\)

3. \(\dfrac{\sin\;2\theta}{\sin\;\theta} ~-~ \dfrac{\cos\;2\theta}{\cos\;\theta} ~=~ \sec\;\theta\)

4. \(\dfrac{\sin\;3\theta}{\sin\;\theta} ~-~ \dfrac{\cos\;3\theta}{\cos\;\theta} ~=~ 2\)

5. \(\tan\;2\theta ~=~ \dfrac{2}{\cot\;\theta \;-\; \tan\;\theta}\)

6. \(\tan\;3\theta ~=~ \dfrac{3\;\tan\;\theta \;-\; \tan^3 \;\theta}{1 \;-\; 3\;\tan^2 \;\theta}\)

7. \(\tan^2 \;\tfrac{1}{2}\theta ~=~ \dfrac{\tan\;\theta \;-\; \sin\;\theta}{\tan\;\theta \;+\; \sin\;\theta}\)

8. \(\dfrac{\cos^2 \;\psi}{\cos^2 \;\theta} ~=~ \dfrac{1 \;+\; \cos\;2\psi}{1 \;+\; \cos\;2\theta}\)

9. Some trigonometry textbooks used to claim incorrectly that \(\;\sin\;\theta ~+~ \cos\;\theta ~=~ \sqrt{1 \;+\; \sin\;2\theta}\) was an identity. Give an example of a specific angle \(\theta\) that would make that equation false. Is \(\;\sin\;\theta ~+~ \cos\;\theta ~=~ \pm\;\sqrt{1 \;+\; \sin\;2\theta}\) an identity? Justify your answer.

10. Fill out the rest of the table below for the angles \(0^\circ < \theta < 720^\circ\) in increments of \(90^\circ\), showing \(\theta\), \(\tfrac{1}{2}\theta\), and the signs ($+$ or $-$) of \(\sin\;\theta\) and \(\tan\;\tfrac{1}{2}\theta\).

    \(\theta\)	\(\tfrac{1}{2}\theta\)	\(\sin\;\theta\)	\(\tan\;\tfrac{1}{2}\theta\)
    \(0^\circ - 90^\circ\)	\(0^\circ - 45^\circ\)
    \(90^\circ - 180^\circ\)	\(45^\circ - 90^\circ\)
    \(180^\circ - 270^\circ\)	\(90^\circ - 135^\circ\)
    \(270^\circ - 360^\circ\)	\(135^\circ - 180^\circ\)
    \(360^\circ - 450^\circ\)	\(180^\circ - 225^\circ\)
    \(450^\circ - 540^\circ\)	\(225^\circ - 270^\circ\)
    \(540^\circ - 630^\circ\)	\(270^\circ - 315^\circ\)
    \(630^\circ - 720^\circ\)	\(315^\circ - 360^\circ\)

11. In general, what is the largest value that \(\;\sin\;\theta~\cos\;\theta\;\) can take? Justify your answer.

For Exercises 12- 17, prove the given identity for any right triangle \(\triangle\,ABC$ with $C=90^\circ\).

12. \(\sin\;(A-B) ~=~ \cos\;2B\)

13. \(\cos\;(A-B) ~=~ \sin\;2A\)

14. \(\sin\;2A ~=~ \dfrac{2\;ab}{c^2}\)

15. \(\cos\;2A ~=~ \dfrac{b^2 - a^2}{c^2}\)

16. \(\tan\;2A ~=~ \dfrac{2\;ab}{b^2 - a^2}\)

17. \(\tan\;\tfrac{1}{2}A ~=~ \dfrac{c - b}{a} ~=~ \dfrac{a}{c + b}\)

18. Continuing Exercise 20 from Section 3.1, it can be shown that

\[\begin{split}\begin{align*} r\;(1 \;-\; \cos\;\theta) ~&=~ a\;(1 \;+\; \epsilon)\,(1 \;-\; \cos\;\psi) ~,~\text{and}\\ r\;(1 \;+\; \cos\;\theta) ~&=~ a\;(1 \;-\; \epsilon)\,(1 \;+\; \cos\;\psi) ~, \end{align*}\end{split}\]

where \(\theta\) and \(\psi\) are always in the same quadrant. Show that \(\;\tan\;\tfrac{1}{2}\theta ~=~ \sqrt{\frac{1 \;+\; \epsilon}{1 \;-\; \epsilon}}~ \tan\;\tfrac{1}{2}\psi\;\).

4. 其他恒等式

Other Identities

Though the identities in this section fall under the category of “other”, they are perhaps (along with \(\cos^2 \;\theta + \sin^2 \;\theta = 1\)) the most widely used identities in practice. It is very common to encounter terms such as \(\;\sin\;A + \sin\;B\;\) or \(\;\sin\;A~\cos\;B\;\) in calculations, so we will now derive identities for those expressions. First, we have what are often called the product-to-sum formulas:

(37)\[\sin\;A~\cos\;B ~=~ \phantom{-}\tfrac{1}{2}\;(\sin\;(A+B) ~+~ \sin\;(A-B))\]

(38)\[\cos\;A~\sin\;B ~=~ \phantom{-}\tfrac{1}{2}\;(\sin\;(A+B) ~-~ \sin\;(A-B))\]

(39)\[\cos\;A~\cos\;B ~=~ \phantom{-}\tfrac{1}{2}\;(\cos\;(A+B) ~+~ \cos\;(A-B))\]

(40)\[\sin\;A~\sin\;B ~=~ -\tfrac{1}{2}\;(\cos\;(A+B) ~-~ \cos\;(A-B))\]

We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We see that

\[\begin{split}\begin{align*} \sin\;(A+B) ~+~ \sin\;(A-B) ~&=~ (\sin\;A~\cos\;B ~+~ \cancel{\cos\;A~\sin\;B}) ~+~ (\sin\;A~\cos\;B ~-~ \cancel{\cos\;A~\sin\;B})\\ &=~ 2\;\sin\;A~\cos\;B ~, \end{align*}\end{split}\]

so formula (37) follows upon dividing both sides by $2$. Notice how in each of the above identities a product (e.g. \(\sin\;A~\cos\;B\)) of trigonometric functions is shown to be equivalent to a sum (e.g. \(\tfrac{1}{2}\;(\sin\;(A+B) ~+~ \sin\;(A-B))\)) of such functions. We can go in the opposite direction, with the sum-to-product formulas:

(41)\[\sin\;A ~+~ \sin\;B ~=~ \phantom{-}2\;\sin\;\tfrac{1}{2}(A+B)~ \cos\;\tfrac{1}{2}(A-B)\]

(42)\[\sin\;A ~-~ \sin\;B ~=~ \phantom{-}2\;\cos\;\tfrac{1}{2}(A+B)~ \sin\;\tfrac{1}{2}(A-B)\]

(43)\[\cos\;A ~+~ \cos\;B ~=~ \phantom{-}2\;\cos\;\tfrac{1}{2}(A+B)~ \cos\;\tfrac{1}{2}(A-B)\]

(44)\[\cos\;A ~-~ \cos\;B ~=~ -2\;\sin\;\tfrac{1}{2}(A+B)~\sin\;\tfrac{1}{2}(A-B)\]

These formulas are just the product-to-sum formulas rewritten by using some clever substitutions: let \(x=\frac{1}{2}(A+B)\) and \(y=\frac{1}{2}(A-B)\). Then $x+y=A$ and $x-y=B$. For example, to derive formula 3.43, make the above substitutions in formula (39) to get

\[\begin{split}\begin{align*} \cos\;A ~+~ \cos\;B ~&=~ \cos\;(x+y) ~+~ \cos\;(x-y)\\ &=~ 2\;\cdot\;\tfrac{1}{2}(\cos\;(x+y) ~+~ \cos\;(x-y))\\ &=~ 2\;\cos\;x~\cos\;y\qquad\qquad\text{(by formula (3.39)}\\ &=~ 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B) ~. \end{align*}\end{split}\]

The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).

Example 3.16

We are now in a position to prove Mollweide’s equations, which we introduced in Section 2.3: For any triangle \(\triangle\,ABC\),

\[\frac{a-b}{c} ~=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C} \qquad\text{and}\qquad \frac{a+b}{c} ~=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C} ~.\]

First, since \(C=2\;\cdot\;\tfrac{1}{2}C\), by the double-angle formula we have \(\;\sin\;C = 2\;\sin\;\tfrac{1}{2}C~\cos\;\tfrac{1}{2}C\). Thus,

\[\begin{split}\begin{align*} \frac{a-b}{c} ~&=~ \frac{a}{c} ~-~ \frac{b}{c} ~=~ \frac{\sin\;A}{\sin\;C} ~-~ \frac{\sin\;B}{\sin\;C}\quad\text{(by the Law of Sines)}\\ &=~ \frac{\sin\;A ~-~ \sin\;B}{\sin\;C} ~=~ \frac{\sin\;A ~-~ \sin\;B}{2\;\sin\;\tfrac{1}{2}C~\cos\;\tfrac{1}{2}C}\\ &=~ \frac{2\;\cos\;\tfrac{1}{2}(A+B)~\sin\;\tfrac{1}{2}(A-B)}{2\;\sin\;\tfrac{1}{2}C~ \cos\;\tfrac{1}{2}C}\quad\text{(by formula (3.42))}\\ &=~ \frac{\cos\;\tfrac{1}{2}(180^\circ - C)~\sin\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C~ \cos\;\tfrac{1}{2}C}\quad\text{(since $A+B=180^\circ - C$)}\\ &=~ \frac{\cancel{\cos\;(90^\circ - \tfrac{1}{2}C)}~\sin\;\tfrac{1}{2}(A-B)}{ \cancel{\sin\;\tfrac{1}{2}C}~\cos\;\tfrac{1}{2}C}\\ &=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C}\quad\text{(since $\;\cos\;(90^\circ - \tfrac{1}{2}C) = \sin\;\tfrac{1}{2}C$)}~. \end{align*}\end{split}\]

This proves the first equation. The proof of the other equation is similar (see Exercise 7).

Example 3.17

Using Mollweide’s equations, we can prove the Law of Tangents: For any triangle $triangle,ABC$,

\[\frac{a-b}{a+b} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~,\quad \frac{b-c}{b+c} ~=~ \frac{\tan\;\frac{1}{2}(B-C)}{\tan\;\frac{1}{2}(B+C)} ~,\quad \frac{c-a}{c+a} ~=~ \frac{\tan\;\frac{1}{2}(C-A)}{\tan\;\frac{1}{2}(C+A)} ~.\]

We need only prove the first equation; the other two are obtained by cycling through the letters. We see that

\[\begin{split}\begin{align*} \frac{a-b}{a+b} ~&=~ \dfrac{\dfrac{a-b}{c}}{\dfrac{a+b}{c}} ~=~ \dfrac{\dfrac{\sin\;\tfrac{1}{2}(A-B)}{\cos\;\tfrac{1}{2}C}}{ \dfrac{\cos\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C}}\quad\text{(by Mollweide's equations)}\\ &=~ \dfrac{\sin\;\tfrac{1}{2}(A-B)}{\cos\;\tfrac{1}{2}(A-B)} \;\cdot\; \dfrac{\sin\;\tfrac{1}{2}C}{\cos\;\tfrac{1}{2}C}\\ &=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;\tfrac{1}{2}C ~=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \tan\;(90^\circ - \tfrac{1}{2}(A+B)) \quad\text{(since $C=180^\circ - (A+B)$)}\\ &=~ \tan\;\tfrac{1}{2}(A-B) \;\cdot\; \cot\;\tfrac{1}{2}(A+B)\quad\text{(since $\tan\;(90^\circ - \tfrac{1}{2}(A+B)) = \cot\;\tfrac{1}{2}(A+B)$, see Section 1.5)}\\ &=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~.\quad \end{align*}\end{split}\]

[qed]

Example 3.18

For any triangle \(\triangle\,ABC\), show that

\[\cos\;A ~+~ \cos\;B ~+~ \cos\;C ~=~ 1 ~+~ 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~.\]

Solution: Since \(\;\cos\;(A+B+C) = \cos\;180^\circ = -1\), we can rewrite the left side as

\[\begin{split}\begin{align*} \cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~ 1 \;+\; (\cos\;(A+B+C) \;+\; \cos\;C) \;+\; (\cos\;A \;+\; \cos\;B)~~\text{, so by formula (3.43)}\\ &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B+2C)~\cos\;\tfrac{1}{2}(A+B) \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)~\cos\;\tfrac{1}{2}(A-B)\\ &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)~\left( \cos\;\tfrac{1}{2}(A+B+2C) \;+\; \cos\;\tfrac{1}{2}(A-B) \right) ~~\text{, so}\\ &=~ 1 \;+\; 2\;\cos\;\tfrac{1}{2}(A+B)\;\cdot\;2\;\cos\;\tfrac{1}{2}(A+C)~ \cos\;\tfrac{1}{2}(B+C)~~\text{by formula (3.43),}\\ \end{align*}\end{split}\]

since \(\tfrac{1}{2}\left( \tfrac{1}{2}(A+B+2C) + \tfrac{1}{2}(A-B) \right) = \tfrac{1}{2}(A+C)\) and \(\tfrac{1}{2}\left( \tfrac{1}{2}(A+B+2C) - \tfrac{1}{2}(A-B) \right) = \tfrac{1}{2}(B+C)\). Thus,

\[\begin{split}\begin{align*} \cos\;A \;+\; \cos\;B \;+\; \cos\;C ~&=~ 1 \;+\; 4\;\cos\;(90^\circ - \tfrac{1}{2}C)~\cos\;(90^\circ - \tfrac{1}{2}B)~ \cos\;(90^\circ - \tfrac{1}{2}A)\\ &=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}C~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}A ~~,\text{ so rearranging the order gives}\\ &=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~. \end{align*}\end{split}\]

Example 3.19

For any triangle \(\triangle\,ABC\), show that \(\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~ \sin\;\tfrac{1}{2}C \;\le\; \frac{1}{8}\;\).

Solution: Let \(u=\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C\). Apply formula (40) to the first two terms in $u$ to get

\[u ~=~ -\tfrac{1}{2}\;(\cos\;\tfrac{1}{2}(A+B) \;-\; \cos\;\tfrac{1}{2}(A-B))~ \sin\;\tfrac{1}{2}C ~=~ \tfrac{1}{2}\;(\cos\;\tfrac{1}{2}(A-B) \;-\; \cos\;\tfrac{1}{2}(A+B))~\cos\;\tfrac{1}{2}(A+B) ~,\]

since \(\;\sin\;\tfrac{1}{2}C = \cos\;\tfrac{1}{2}(A+B)\), as we saw in Example 3.18. Multiply both sides by $2$ to get

\[\cos^2 \;\tfrac{1}{2}(A+B) ~-~ \cos\;\tfrac{1}{2}(A-B)~\cos\;\tfrac{1}{2}(A+B) ~+~ 2u ~=~ 0 ~,\]

after rearranging the terms. Notice that the expression above is a quadratic equation in the term \(\;\cos\;\tfrac{1}{2}(A+B)\). So by the quadratic formula,

\[\cos\;\tfrac{1}{2}(A+B) ~=~ \frac{\cos\;\tfrac{1}{2}(A-B) \;\pm\; \sqrt{\cos^2 \;\tfrac{1}{2}(A-B) - 4(1)(2u)}}{2} ~~,\]

which has a real solution only if the quantity inside the square root is nonnegative. But we know that \(\;\cos\;\tfrac{1}{2}(A+B)\;\) is a real number (and, hence, a solution exists), so we must have

\[\cos^2 \;\tfrac{1}{2}(A-B) \;- \; 8u ~\ge~ 0 \quad\Rightarrow\quad u ~\le~ \tfrac{1}{8}\; \cos^2 \;\tfrac{1}{2}(A-B) ~\le~ \tfrac{1}{8} \quad\Rightarrow\quad \;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~\le~ \tfrac{1}{8} ~.\]

Example 3.20

For any triangle \(\triangle\,ABC\), show that \(\;1 ~<~ \cos\;A + \cos\;B + \cos\;C ~\le~ \tfrac{3}{2}\;\).

Solution: Since \(0^\circ < A,\; B,\; C < 180^\circ\), the sines of \(\tfrac{1}{2}A\), \(\tfrac{1}{2}B\), and \(\tfrac{1}{2}C\) are all positive, so

\[\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~ > ~ 1\]

by Example 3.18. Also, by Examples 3.18 and 3.19 we have

\[\cos\;A \;+\; \cos\;B \;+\; \cos\;C ~=~ 1 \;+\; 4\;\sin\;\tfrac{1}{2}A~\sin\;\tfrac{1}{2}B~\sin\;\tfrac{1}{2}C ~\le~ 1 \;+\; 4\;\cdot\;\tfrac{1}{8} ~=~ \tfrac{3}{2} ~.\]

Hence, \(\;1 ~<~ \cos\;A + \cos\;B + \cos\;C ~\le~ \tfrac{3}{2}\;\).

Example 3.21

Recall Snell’s law from Example 3.12 in Section 3.2: \(n_1 ~\sin\;\theta_1 = n_2 ~\sin\;\theta_2\). Use it to show that the p-polarization transmission Fresnel coefficient defined by

(45)\[t_{1\;2\;p} ~=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2}\]

can be written as:

\[t_{1\;2\;p} ~=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~ \cos\;(\theta_1 - \theta_2)} ~.\]

Solution: Multiply the top and bottom of \(t_{1\;2\;p}\) by \(\;\sin\;\theta_1 ~\sin\;\theta_2\;\) to get:

\[\begin{split}\begin{align*} t_{1\;2\;p} ~&=~ \frac{2\;n_1 ~\cos\;\theta_1}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2} \;\cdot\; \frac{\sin\;\theta_1 ~ \sin\;\theta_2}{\sin\;\theta_1 ~ \sin\;\theta_2}\\[7pt] &=~ \frac{2\;(n_1 ~\sin\;\theta_1)~\cos\;\theta_1 ~\sin\;\theta_2}{ (n_2 ~\sin\;\theta_2)~\sin\;\theta_1 ~\cos\;\theta_1 ~+~ (n_1 ~\sin\;\theta_1)~\sin\;\theta_2 ~\cos\;\theta_2}\\[7pt] &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \sin\;\theta_1 ~\cos\;\theta_1 ~+~ \sin\;\theta_2 ~\cos\;\theta_2} \qquad\text{(by Snell's law)}\\[7pt] &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \tfrac{1}{2}\;(\sin\;2\,\theta_1 ~+~ \sin\;2\theta_2)} \qquad\text{(by the double-angle formula)}\\[7pt] &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{ \tfrac{1}{2}\;(2\;\sin\;\tfrac{1}{2}(2\theta_1 + 2\theta_2)~ \cos\;\tfrac{1}{2}(2\theta_1 - 2\theta_2))} \qquad\text{(by formula (3.41))}\\[7pt] &=~ \frac{2\;\cos\;\theta_1~\sin\;\theta_2}{\sin\;(\theta_1 + \theta_2)~ \cos\;(\theta_1 - \theta_2)} \end{align*}\end{split}\]

Example 3.22

In an AC electrical circuit, the instantaneous power $p(t)$ delivered to the entire circuit in the sinusoidal steady state at time $t$ is given by

\[p(t) ~=~ v(t)\;i(t) ~,\]

where the voltage $v(t)$ and current $i(t)$ are given by

\[\begin{split}\begin{align*} v(t) ~&=~ V_m \;\cos\;\omega t ~,\\ i(t) ~&=~ I_m \;\cos\;(\omega t + \phi)~, \end{align*}\end{split}\]

for some constants $V_m$, $I_m$, $omega$, and $phi$. Show that the instantaneous power can be written as

\[p(t) ~=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;\phi ~+~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \phi) ~.\]

Solution: By definition of $p(t)$, we have

\[\begin{split}\begin{align*} p(t) ~&=~ V_m \;I_m \;\cos\;\omega t~\cos\;(\omega t + \phi)\\ &=~ V_m \;I_m \;\cdot\;\tfrac{1}{2}(\cos\;(2\omega t + \phi) \;+\; \cos\;(-\phi)) \qquad&&\text{(by formula (3.39))}\\ &=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;\phi ~+~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \phi) \qquad&&\text{(since $\cos\;(-\phi) = \cos\;\phi$)}~. \end{align*}\end{split}\]

练习

Exercises

1. Prove formula (38).

2. Prove formula (39).

3. Prove formula (40).

4. Prove formula (41).

5. Prove formula (42).

6. Prove formula (44).

7. Prove Mollweide’s second equation: For any triangle \(\triangle\,ABC\), \(~\dfrac{a+b}{c} ~=~ \dfrac{\cos\;\tfrac{1}{2}(A-B)}{\sin\;\tfrac{1}{2}C}~\).

8. Continuing Example 3.21, use Snell’s law to show that the p-polarization reflection Fresnel coefficient

   \[r_{1\;2\;p} ~=~ \frac{n_2 ~\cos\;\theta_1 ~-~ n_1 ~\cos\;\theta_2}{n_2 ~\cos\;\theta_1 ~+~ n_1 ~\cos\;\theta_2}\]

   can be written as:

   \[r_{1\;2\;p} ~=~ \frac{\tan\;(\theta_1 - \theta_2)}{\tan\;(\theta_1 + \theta_2)}\]

9. There is a more general form for the instantaneous power \(p(t) = v(t)\;i(t)\) in an electrical circuit than the one in Example 3.22. The voltage $v(t)$ and current $i(t)$ can be given by

   \[\begin{split}\begin{align*} v(t) ~&=~ V_m \;\cos\;(\omega t + \theta)~,\\ i(t) ~&=~ I_m \;\cos\;(\omega t + \phi)~, \end{align*}\end{split}\]

   where \(\theta\) is called the phase angle. [4] Show that $p(t)$ can be written as

   \[p(t) ~=~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(\theta - \phi) ~+~ \tfrac{1}{2}\,V_m \; I_m \;\cos\;(2\omega t + \theta + \phi) ~.\]
   [4]

   Though it does not matter for this exercise, none of the angles in these formulas are measured in degrees. We will discuss their unit of measurement in Chapter 4.

   For Exercises 10-15, prove the given identity or inequality for any triangle \(\triangle\,ABC\).

10. \(\sin\;A \;+\; \sin\;B \;+\; \sin\;C ~=~ 4\;\cos\;\tfrac{1}{2}A~\cos\;\tfrac{1}{2}B~\cos\;\tfrac{1}{2}C\) (Hint: Mimic Example 3.18 using \((\sin\;A \;+\; \sin\;B) \;+\; (\sin\;C \;-\; \sin\;(A+B+C))\).)

11. \(\cos\;A \;+\; \cos\;(B-C) ~=~ 2\;\sin\;B~\sin\;C\)

12. \(\sin\;2A \;+\; \sin\;2B \;+\; \sin\;2C ~=~ 4\;\sin\;A~\sin\;B~\sin\;C\) (Hints: Group \(\sin\;2B\) and $sin;2C$ together, use the double-angle formula for \(\sin\;2A\), use Exercise 11.)

13. \(\dfrac{a-b}{a+b} ~=~ \dfrac{\sin\;A \;-\; \sin\;B}{\sin\;A \;+\; \sin\;B}\)

14. \(\cos\;\tfrac{1}{2}A ~=~ \sqrt{\dfrac{s\;(s-a)}{bc}}~~\) and \(~~\sin\;\tfrac{1}{2}A ~=~ \sqrt{\dfrac{(s-b)\;(s-c)}{bc}}\;\), where \(s=\tfrac{1}{2}(a+b+c)\) (Hint: Use the Law of Cosines to show that \(2bc\;(1 + \cos\;A) ~=~ 4s\;(s-a)\).)

15. \(\tfrac{1}{2}\;(\sin\;A \;+\; \sin\;B) ~\le~ \sin\;\tfrac{1}{2}(A+B)\) (Hint: Show that \(\sin\;\tfrac{1}{2}(A+B) \;-\; \tfrac{1}{2}\;(\sin\;A \;+\; \sin\;B) \;\ge\; 0\).)

16. In Example 3.20, which angles $A$, $B$, $C$ give the maximum value of \(\cos\;A \;+\; \cos\;B \;+\; \cos\;C\;\)?