So far we know a few relations between the trigonometric functions. For example, we know the
reciprocal relations:
\(\csc\;\theta ~=~ \dfrac{1}{\sin\;\theta}\qquad\) when \(\sin\;\theta \ne 0\)
\(\sec\;\theta ~=~ \dfrac{1}{\cos\;\theta}\qquad\) when \(\cos\;\theta \ne 0\)
\(\cot\;\theta ~=~ \dfrac{1}{\tan\;\theta}\qquad\) when \(\tan\;\theta\) is defined and not $0$
\(\sin\;\theta ~=~ \dfrac{1}{\csc\;\theta}\qquad\) when \(\csc\;\theta\) is defined and not $0$
\(\cos\;\theta ~=~ \dfrac{1}{\sec\;\theta}\qquad\) when \(\sec\;\theta\) is defined and not $0$
\(\tan\;\theta ~=~ \dfrac{1}{\cot\;\theta}\qquad\) 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:
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
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:
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
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:
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:
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
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
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
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:
By (5) both sides of the last equation are indeed equal. Thus, the original identity holds.
Example 3.7
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 $a$:
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.
We showed that \(\;\sin\;\theta ~=~ \pm\,\sqrt{1 ~-~ \cos^2 \;\theta}\;\) for all \(\theta\). Give an example of an angle \(\theta\) such that \(\sin\;\theta ~=~ -\sqrt{1 ~-~ \cos^2 \;\theta}\;\).
We showed that \(\;\cos\;\theta ~=~ \pm\,\sqrt{1 ~-~ \sin^2 \;\theta}\;\) for all \(\theta\). Give an example of an angle \(\theta\) such that \(\cos\;\theta ~=~ -\sqrt{1 ~-~ \sin^2 \;\theta}\;\).
Suppose that you are given a system of two equations of the following form: [1]
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?
Similar to Exercise 16 , find an expression for \(\cos\;\theta\) solely in terms of \(\tan\;\theta\).
Find an expression for \(\tan\;\theta\) solely in terms of \(\sin\;\theta\), and one solely in terms of \(\cos\;\theta\)
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.)
Show that each trigonometric function can be put in terms of the sine function.
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:
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
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 \(90^\circ\):
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
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 :
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:
This, combined with replacing $B$ by $-B$ and using the relation \(\tan\;(-\theta) = -\tan\;\theta\), gives us the addition and subtraction formulas for tangent:
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:
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:
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
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
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
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.
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 \(\theta\).
Continuing Example 3.12, use Snell’s law to show that the s-polarization transmission Fresnel coefficient
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
Use Exercise 3.12 to find the angle between the lines $y=2x+3$ and $y=-5x-4$.
For any triangle \(\triangle\,ABC\), show that \(\;\cot\;A~\cot\;B ~+~ \cot\;B~\cot\;C ~+~ \cot\;C~\cot\;A ~=~ 1\). (Hint: Use Exercise9 and \(C=180^\circ - (A+B)\).)
For any positive angles $A$, $B$, and $C$ such that \(A+B+C=90^\circ\), show that
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 $B$.
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 \(\beta\).
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:
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]
Closely related to the double-angle formulas are the half-angle formulas:
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:
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
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:
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.
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 \(\tan\;\tfrac{1}{2}\theta\).
\(\theta\)
\(\tfrac{1}{2}\theta\)
\(\sin\;\theta\)
\(\tan\;\tfrac{1}{2}\theta\)
\(0^\circ - 90^\circ\)
\(0^\circ - 45^\circ\)
\(90^\circ - 180^\circ\)
\(45^\circ - 90^\circ\)
\(180^\circ - 270^\circ\)
\(90^\circ - 135^\circ\)
\(270^\circ - 360^\circ\)
\(135^\circ - 180^\circ\)
\(360^\circ - 450^\circ\)
\(180^\circ - 225^\circ\)
\(450^\circ - 540^\circ\)
\(225^\circ - 270^\circ\)
\(540^\circ - 630^\circ\)
\(270^\circ - 315^\circ\)
\(630^\circ - 720^\circ\)
\(315^\circ - 360^\circ\)
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 \(\triangle\,ABC$ with $C=90^\circ\).
where \(\theta\) and \(\psi\) are always in the same quadrant. Show that \(\;\tan\;\tfrac{1}{2}\theta ~=~ \sqrt{\frac{1 \;+\; \epsilon}{1 \;-\; \epsilon}}~ \tan\;\tfrac{1}{2}\psi\;\).
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:
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:
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
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,
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
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
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
\(\sin\;2A \;+\; \sin\;2B \;+\; \sin\;2C ~=~ 4\;\sin\;A~\sin\;B~\sin\;C\) (Hints: Group\(\sin\;2B\)and $sin;2C$ together, use the double-angle formula for\(\sin\;2A\), use Exercise 11.)
\(\cos\;\tfrac{1}{2}A ~=~ \sqrt{\dfrac{s\;(s-a)}{bc}}~~\) 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\(2bc\;(1 + \cos\;A) ~=~ 4s\;(s-a)\).)