Chapter 13
Identities, Inverses, Equations
Sum Identities
Cogito ergo sum.
- Descartes
You’ve learned many trigonometric identities, but before you can be considered fluent in trigonometry, you must learn a few more: the sum identities and double-angle identities for sine and cosine.
Sum Identity for Sine. The following identity holds for all \(\alpha\) and \(\beta\):
\[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha. \]
Proof. For now, I’m going to prove this identity only in the special case when $ $ and $ $ are acute.
On the next page, I’ll explain why the result holds for all angles.
Draw a line. From any point on it, erect a perpendicular 1 unit long. From its top, draw rays making angles $ $ and $ $ with the perpendicular. These will eventually hit the original line, producing the figure at right, whose area we’ll now compute in two ways.
First, by the SAS area formula, its area is
\[ \frac{1}{2}\sec\alpha\sec\beta\sin(\alpha+\beta). \]
Second, applying the SAS area formula to each right triangle and adding the results yields
\[ \frac{1}{2}\sec\alpha\sin\alpha+\frac{1}{2}\sec\beta\sin\beta. \]
If you equate these two expressions, isolate $ (+ ) $ , and simplify, you’ll find that
\[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha, \]
as claimed.
Notice that proof’s cleverness. The sum identity says nothing about area, yet we used areas to prove it. $ ^{*} $ Bringing angle $ (+ ) $ into play and then wrapping it up in a sine by means of the SAS area formula was a particularly fine setup for the old “compute something two ways” trick.
A minor consequence of this sum identity is that we can now obtain a few more exact values of sine.
Example. Find the exact value of $ (75^{}) $ .
\[ \begin{aligned}Solution.\sin(75^{\circ})&=\sin(30^{\circ}+45^{\circ})\\&=\sin(30^{\circ})\cos(45^{\circ})+\sin(45^{\circ})\cos(30^{\circ})\quad(sum identity for sine)\\&=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{2}+\sqrt{6}}{4}.\end{aligned} \]
Next, we’ll establish the sum identity for cosine.
Sum Identity for Cosine. The following identity holds for all \(\alpha\) and \(\beta\):
\[ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta. \]
Proof. As with the previous identity, I’ll first prove this in the special case when $ $ and $ $ are acute. (I’ll then explain how both identities can be extended to all angles.)
Begin with two radii of the unit circle making angles $ $ and $ $ with the positive x-axis as in the figure. We’ll compute AB twice: first with the distance formula, then with the law of cosines.
For the distance formula, we will need $ A^{} $ s coordinates, $ (,) $ , and $ B^{} $ s, $ ((-),(-)) $ , which we can also write as $ (,-) $ , thanks to cosine’s evenness and sine’s oddness. So, by the distance formula,
\[ \begin{aligned}AB&=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}\\&=\sqrt{(\cos\alpha-\cos\beta)^{2}+(\sin\alpha+\sin\beta)^{2}}\\&=\sqrt{2+2\sin\alpha\sin\beta-2\cos\alpha\cos\beta}\quad&(by the Pythagorean identity).\end{aligned} \]
On the other hand, the law of cosines tells us that
\[ AB=\sqrt{2-2\cos(\alpha+\beta)}. \]
Equating the two expressions for AB, squaring, and then simplifying yields the sum identity.
We’ve proved that the sum identities hold when $ $ and $ $ are acute. Now let’s see why they always hold. The explanation hinges on two simple observations. First, the identities still hold if either $ $ or $ $ is zero. $ ^{*} $ Second, we can think of any angle whatsoever as a multiple of $ 90^{} $ plus some acute angle (or zero). [Examples: $ 328{}=3(90{})+58^{} $ ; $ -838{}=-10(90{})+62^{} $ .]
A completely rigorous proof of sine’s sum identity would require a tedious number of cases, but we can expose the proof’s central idea – which is what matters – by picking two essentially arbitrary angles $ (328^{} $ and $ -838^{}) $ and showing why sine’s sum identity holds for those angles. The argument we’ll use will work (with a few minor tweaks) for any two angles, which is why the identity holds universally.
\[ \mathsf{C l a i m.}\sin\left(328^{\circ}+(-838^{\circ})\right)=(\sin328^{\circ})(\cos(-838^{\circ}))+(\sin(-838^{\circ}))(\cos328^{\circ}). \]
Proof. By first sifting out the “acute parts” of the angles, as explained above, and then bundling up the leftover right angles, we can rewrite the expression on the left-hand side as follows:
\[ \begin{aligned}\sin\big(328^{\circ}+(-838^{\circ})\big)&=\sin\lbrack(3(90^{\circ})+\mathbf{58^{\circ}})+(-10(90^{\circ})+\mathbf{62^{\circ}})]\\&=\sin\lbrack(\mathbf{58^{\circ}}+\mathbf{62^{\circ}})-7(90^{\circ})\rbrack.\end{aligned} \]
Thinking about the unit circle for a few moments should convince you that for any angle $ $ , we have $ (- 7(90^{})) = $ . Consequently, we can rewrite the last expression above as
\[ cos(58^{\circ}+62^{\circ}). \]
Expanding this with cosine’s sum identity (which we’ve proved for acute angles like these!) yields
\[ \cos58^{\circ}\cos62^{\circ}-\sin58^{\circ}\sin62^{\circ}. \]
To inject our original angles back into the scene, we’ll rewrite this expression in an ungainly form:
\[ \cos\big(328^{\circ}-3(90^{\circ})\big)\cos\big(-838^{\circ}+10(90^{\circ})\big)-\sin\big(328^{\circ}-3(90^{\circ})\big)\sin\big(-838^{\circ}+10(90^{\circ})\big). \]
Thinking again about the unit-circle definitions of sine and cosine, we see that this is equal to
\[ \cos(328^{\circ}+90^{\circ})\cos(-838^{\circ}+180^{\circ})-\sin(328^{\circ}+90^{\circ})\sin(-838^{\circ}+180^{\circ}). \]
More unit-circle contemplation (of the sort you did in Chapter 10, Exercises 41–43) reduces this to
\[ (-\sin328^{\circ})(-\cos(-838^{\circ}))-(\cos328^{\circ})(-\sin(-838^{\circ})). \]
Finally, sorting out the negatives reveals that this is equal to
\[ (\sin328^{\circ})(\cos(-838^{\circ}))+(\sin(-838^{\circ}))(\cos328^{\circ}),\mathrm{as\ claimed}. \]
The same sort of argument will justify cosine’s sum identity. Of course, there’s no need to go through this complicated procedure in practice. You should, however, do it at least once (in exercise 7) to convince yourself that the sum identities do indeed hold for all angles. If nothing else, this is excellent practice in thinking about the unit-circle definitions of sine and cosine.
Exercises
Using the equation $ 15^{} = 45^{} + (-30^{}) $ , find the exact value of $ ^{} $
Find exact values for sin 105°, cos 15°, cos 75°, cos 105°, tan 15°, tan 75°, and tan 105°.
Derive sine’s difference identity (i.e. an identity for $ (- ) $ ). [Hint: Revisit the trick in Exercise 1.]
Derive cosine’s difference identity.
Simplify as much as possible:
\[ \begin{aligned}a)\frac{\sin(\theta+\phi)-\sin\theta\cos\phi}{\cos\theta\cos\phi}\qquad&b)(-\sin\theta\sin\phi+\cos\theta\cos\phi)^{2}+\sin^{2}(\theta+\phi)\qquad c)\frac{\sin\gamma\cos\delta-\sin\delta\cos\gamma}{\cos(\gamma-\delta)}\end{aligned} \]
Derive an identity for $ (+ ) $ as follows: rewrite this expression in terms of sine and cosine, then apply the sum identities you already know, and finally, divide the top and bottom of the resulting fraction by $ $ .
Just once, for the good of your soul, use the sort of argument in the “Claim” above to show that cosine’s sum identity really does hold when $ = 133^{} $ and $ = 204^{} $ .
Use the sum identities to reconfirm the following identities (which you can also read off of the unit circle):
\[ \mathsf{a})\sin(\pi/2+\theta)=\cos\theta \]
\[ \mathsf{b})\cos(\theta+\pi)=-\cos\theta \]
\[ c)\sin(\pi-\theta)=\sin\theta \]
$ (-)=$
$ (+ 2k) = $ for all integers k.
- Express sin( $ + + $ ) in terms of the sines and cosines of $ $ , $ $ , and $ $ .
Double-Angle Identities
The double-angle angle identities follow directly from the sum identities:
Double Angle Identities. The following identities hold for all \(\theta\):
\[ \sin(2\theta)=2\sin\theta\cos\theta \]
\[ \cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta \]
Proofs. Rewrite $ 2$ as $ (+ ) $ and apply each sum identity. I leave it to you to verify the details.
Sine’s double-angle identity is particularly easy to remember because it sounds like a line of poetry; it scans in perfect trochaic tetrameter:
\[ \stackrel{/}{\mathrm{Two~sine~}}\mid\stackrel{/}{\mathrm{theta~}}\mid\stackrel{/}{\mathrm{cosine~}}\mid\stackrel{/}{\mathrm{theta}} \]
Memorize both of these double-angle identities. They are often handy in calculus.
Exercises
Simplify as much as possible: $ {2}+(){2} $
Express sin(3θ) in terms of sin θ. [Hints: 3 = 2 + 1. Sum identity, Pythagorean identity.]
Prove that sin(4θ) = 4 sin θ cos³ θ − 4 sin³ θ cos θ. [Hint: 4 = 2 · 2. Double-angle identity.]
Express cos(3θ) in terms of cos θ.
Express cos(4θ) in terms of sin θ.
Verify that sine’s double angle identity is correct when $ = 30^{} $ .
Verify that cosine’s double angle identity is correct when $ = 7/6 $ .
Prove that the following identities hold for all $ $ :
- (cos θ − sin θ)² = 1 − sin(2θ)
\[ \cos(2\theta)=1-2\sin^{2}\theta \]
\[ \mathsf{c})\cos(2\theta)=2\cos^{2}\theta-1 \]
\[ \mathsf{d})\cos^{4}\theta-\sin^{4}\theta=\cos(2\theta) \]
\[ \frac{\sin2\theta}{\sin\theta}-\frac{\cos2\theta}{\cos\theta}=\sec\theta \]
\[ \mathrm{f)}\frac{\cos^{2}\theta}{1+\sin\theta}=1-\sin\theta \]
- $ 1 + ^{2} = ^{2} $ [Hint: Divide both sides of the Pythagorean identity by… something.]
\[ \mathsf{h})1+\cot^{2}\theta=\csc^{2}\theta \]
\[ i)\sin^{4}\theta-\sin^{2}\theta=\cos^{4}\theta-\cos^{2}\theta \]
- The Pythagorean identity and cosine’s double-angle identity both involve $ ^{2}$ and $ ^{2}$ , which suggests that we might fruitfully combine them. By adding their corresponding sides and massaging the results, you can derive an identity that proves quite useful in integral calculus: $ ^{2}= (1 + )/2 $ .
- Derive it. b) Derive a similar identity for $ ^{2}$ by subtracting where you added in Part A.
- Take square roots of both sides of the identities you derived in exercise 18, then substitute $ /2 $ for $ $ . The results are sometimes called half-angle identities for cosine and sine.
Find these two half-angle identities. b) Use them to find exact expressions for sin 15° and cos 15°.
Use the results of Part B to find an exact expression for $ ^{} $ .
What is the greatest possible value of $ $ ? How do you know it is the greatest? What is the smallest positive angle $ $ at which this maximum value is attained?
Sketch graphs of the following functions: a) $ y = 4 x x $ , b) $ y = ^{4} x - ^{4} x $ .
Enisoc (Inverse Cosine)
Until now, we’ve been able to make do with a naïve understanding of the inverse trigonometric functions. For instance, until now, we have understood $ ^{-1}(1/2) $ as meaning “the angle whose cosine is 1/2”. Well, that definite article was valid so long as we were just concerned with solving triangles, because there is indeed a unique “triangle angle” (that is, an angle between 0 and $ $ ) whose cosine is 1/2. $ ^{*} $ However, in a triangle-free context, that naïve definition of inverse cosine ceases to work, because there are infinitely many numbers whose cosine is 1/2. $ ^{} $ Which of those infinitely many candidates is $ ^{-1}(1/2) $ ?
To provide an unambiguous answer, we make the following formal definition:
Definition (Inverse Cosine). cos⁻¹ x is the number in the range [0, π] whose cosine is x. (Note: Some people call inverse cosine “arccosine”, and write arccos x instead of cos⁻¹ x. The meaning, however, is the same.)
This definition makes inverse cosine into a true function, for it now satisfies the “one in, one out” criterion. For any permissible input, it provides a unique output. Yes, there are infinitely many angles whose cosine is 1/2, but only one of them lies in the range $ [0,] $ , namely, $ /3 $ . Thus, $ ^{-1}(1/2) $ equals $ /3 $ , and nothing else. This sort of range restriction may seem peculiar, but you have actually met it before: Each positive number x has two square roots, but the function $ $ specifically picks out the positive root.
Exercises
With the help of the unit circle, explain why there is a unique angle in $ [0,] $ whose cosine is -0.249.
True or false: For every r in cosine’s range, $ [-1, 1] $ , there is a unique angle in $ [0, ] $ whose cosine is r.
Explain why there is a not a unique angle in $ [0, ] $ whose sine is 0.7734.
True or false. (Explain your answer.)
$ ^{-1}(1)=2$ , since $ (2)=1 $ . b) $ ^{-1}(-1) $ is undefined. c) $ ^{-1}(-1/2)=-2/3 $ .
$ ^{-1}(1/2)=/3 $ e) Inverse cosine is an even function. f) Inverse cosine is an odd function.
$ (^{-1}(/2)) = /2 $ h) $ (^{-1}(-.83)) = -0.83 $
\[ \cos(\cos^{-1}(r))=r. \]
\[ \cos^{-1}(\cos\pi)=\pi \]
- $ ^{-1}((2)) = 2$ [Careful.] I) $ ^{-1}() = 2.31 $ m) $ ^{-1}() = 4 $
- Find the exact values of…
$ ^{-1}(1) $ b) $ ^{-1}(-1) $ c) $ ^{-1}(0) $ d) $ ^{-1}(1/2) $ e) $ ^{-1}(/2) $
$ ^{-1}(-1/2) $ g) $ ^{-1}(-/2) $ h) $ ^{-1}((/3)) $ i) $ ^{-1}((/7)) $
$ ^{-1}((5)) $ k) $ ^{-1}((10/9)) $ l) $ ^{-1}((-3/14)) $
arccos(0) n) $ ((e/)) $ o) $ ((/e)) $ p) $ ((39/20)) $
I nverse Sine and Tangent
Exercise 24 shows that to define inverse sine, we’ll need to restrict its range in a slightly different way. While we’re at it, we’ll define inverse tangent as well.
Definitions.
$ ^{-1} x $ is the number in $ [- /2, /2] $ whose sine is x.
$ ^{-1} x $ is the number in $ [- /2, /2] $ whose tangent is x.
Just as you would probably guess, $ ^{-1} x $ and $ ^{-1} x $ are also sometimes written arcsin x and arctan x. With the definitions above in place, you’ll be ready to use the inverse trigonometric functions in calculus, where they occur naturally in triangle-free contexts (as antiderivatives of non-trigonometric functions).
As for inverses of the reciprocal trigonometric functions, one can define them, but no one bothers. They are used so infrequently that there isn’t even a standard convention for restricting their ranges. Nor, for that matter, do they even appear on most calculators.
Exercises
Explain why there is a unique angle in $ [-/2, /2] $ whose sine is 0.7734. [Compare exercise 24.]
True or false: For every r in sine’s range, there is a unique angle in $ $ whose sine is r.
Explain why there is a unique angle in $ [-/2, /2] $ whose tangent is -0.889.
True or false: For every r in tangent’s range, there is a unique angle in $ [- /2, /2] $ whose tangent is r.
True or false. (Explain your answer.)
- $ ^{-1}(1)=5/4 $ , since $ (5/4)=1 $ . b) $ ^{-1}(2) $ is undefined. c) $ ^{-1}(2) $ is undefined. d) $ ^{-1}()=0 $ . e) $ $ . f) $ ^{-1}0=$ . g) $ ^{-1}0=0 $ . h) $ ^{-1}((/4))=/2 $ . i) $ (^{-1}(123456789))=123456789 $ . j) $ ^{-1}((123456789))=123456789 $ .
- Find the exact values of…
- $ ^{-1}(1) $ b) $ (-1) $ c) $ ^{-1}(0) $ d) $ ^{-1}(-1/2) $ e) $ (/2) $ f) $ {-1}({2}(/5)+^{2}(/5)) $ g) $ ^{-1}(-/2) $ h) $ ^{-1}((3)) $ i) $ (-1) $ j) $ ^{-1}((7/2)) $ k) $ ^{-1}((10/9)) $ l) $ ^{-1}((-3/14)) $
Inverse tangent is the only inverse trigonometric function whose graph you are ever likely to encounter. By thinking carefully about this function’s definition, sketch its graph. [Hint: Think. What is the domain of inverse tangent? What does this function do to zero? To large positive inputs? To large negative inputs? Etc.]
Is inverse tangent an odd function? An even function? Neither? Explain your answer.
Is inverse sine an odd function? An even function? Neither? Explain your answer.
If we could also restrict inverse cosine’s range to $ [- /2, /2] $ , all three inverse trig functions would have the same range, and life would be a bit easier. Explain why we can’t do this.
In calculus, you’ll meet this Taylor series expansion: $ ^{-1}x = x - (x^{3}/3) + (x^{5}/5) - (x^{7}/7) + (x^{9}/9) - $ Letting x = 1 yields the mysterious equation
\[ \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\cdots, \]
links $ $ to the odd numbers via an infinite sum. Ponder this.
Solving Trigonometric Equations
Trigonometric functions are periodic, so equations containing them often have infinitely many solutions. To solve such an equation, the trick is to find all solutions that occur in one trip around the unit circle. (There will generally be a finite number of these.) Once you’ve found these, finding the others is easy, thanks to the trig functions’ periodicity. If you visualize everything on the unit circle, all will be well.
Example 1. Solve the equation $ 2 = $ .
Solution. This is equivalent to $ =/2 $ . In one trip around the circle (as $ $ runs from 0 to $ 2$ ), we’ll encounter exactly two solutions to this equation, as the figure makes clear. Of course, we know what these two solutions will be: $ /3 $ and its supplement, $ 2/3 $ .
Each solution, however, corresponds to infinitely many others. For instance, not only is $ /3 $ a solution, but so are $ (/3)+2$ , $ (/3)-2$ , $ (/3)+4$ , $ (/3)-4$ , and in general, $ (/3)+2k $ for any integer value of k. The same thing occurs, naturally, at $ 2/3 $ .
Consequently, our original equation’s full set of solutions is:
$ (/3) + 2k $ for all integers k, and
$ (2/3)+2k $ for all integers k.
Some trigonometric equations require some preliminary algebra, as in the following example.
Example 2. Solve the equation $ 2 ^{2} = $ .
Solution. Being wise algebraists, we do not divide both sides by $ $ , lest we lose precious solutions.* Rather, we put everything on one side and then factor out $ $ to obtain
\[ \cos\phi\left(2\cos\phi-\sqrt{2}\right)=0. \]
To solve this, we must find the values that make each factor on the left equal to zero.
First, let us determine the values that make the first factor, $ $ , equal to zero. In one loop around the circle, $ = 0 $ obviously has two solutions: $ /2 $ and $ 3/2 $ . Each of these begets an infinite family of solutions. By visualizing where they lie on the unit circle, we can describe them all in one package: $ = /2 + k $ for all integers k.
Next, we must determine the values that make the second factor equal to zero. Well, $ 2 - = 0 $ is equivalent to $ = /2 $ , whose solutions in one loop around the circle are, of course, $ /4 $ . Thus, by cosine’s periodicity, we’ve just found infinitely many more solutions to our equation: $ (/4) + 2k $ for all integers k.
Summing up, our original equation’s solutions are
\[ \begin{aligned}&\pi/2+\pi k for all integers k,and\\&\pm\frac{\pi}{4}+2\pi k for all integers k.\\ \end{aligned} \]
We are trying to solve trigonometric equations, trigonometric identities are often helpful, as you’ll see next.
Example 3. Solve $ x = x $ .
Solution.
\[ \tan x=\cos x \]
\[ \frac{\sin x}{\cos x}=\cos x \]
(using a familiar identity)
\[ \sin x=\cos^{2}x \]
\[ \sin x=1-\sin^{2}x \]
(by the Pythagorean identity)
\[ \sin^{2}x+\sin x-1=0, \]
which is a quadratic in disguise. To make the quadratic explicit, let $ u = x $ . Then we have
\[ u^{2}+u-1=0,\quad\text{so}\quad u=\frac{-1\pm\sqrt{5}}{2}. \]
Translating back to x (that is, letting $ u = x $ again), these two values of u become:
\[ \sin x=\frac{-1+\sqrt{5}}{2}\approx0.618\quad and\quad\sin x=\frac{-1-\sqrt{5}}{2}\approx-1.618. \]
The latter has no solution, because -1.6… is out of sine’s range. The former has infinitely many solutions, two for each trip around the circle, as indicated in the figure. The one in the first quadrant is $ ^{-1}((-1+)/2) $ , so by symmetry, the one in the second quadrant must be approximately $ $ . Thus, the (approximate) solutions to our original equation are
\[ 0.666+2\pi k\mathrm{~a n d~} \]
2.475 + 2πk, for all integers k.
When an equation contains a trigonometric function whose argument is itself a function of the unknown for which we’re trying to solve, a substitution can sometimes help, as it does in this next example.
Example 4. Solve cos(5θ) = -1/2.
Solution. Letting \(u = 5\theta\) yields \(\cos u = -1/2\), whose solutions are
\[ u = \left( \frac{2\pi}{3} \right) + 2\pi k \quad and \quad u = \left( \frac{4\pi}{3} \right) + 2\pi k, for all integers k. \]
Translating back to $ $ , these become
\[ 5\theta = \frac{2\pi}{3} + 2\pi k \quad and \quad 5\theta = \frac{4\pi}{3} + 2\pi k, for all integers k. \]
Dividing each equation’s sides by 5 to solve for $ $ , we obtain the original equation’s solutions:
\[ \frac{2\pi}{15}+\frac{2\pi}{5}\boldsymbol{k}\mathrm{~a n d~} \]
$ +k $ , for all integers k.
Study these examples well. Their techniques can be combined and varied.
Exercises
Is $ (/3)-18$ a solution to the equation in example 1? If not, why not? If so, how is it accounted for in the boldface solutions listed at the end of that example?
Fandor and Juve solve the equation $ /2 $ independently and compare their solutions. Fandor worries that his answer, $ =-(/6)+2k $ for all integers k, must be incorrect, since it differs from Juve’s answer, which is $ =(11/6)+2k $ for all integers k, and Juve is never wrong. Is Fandor’s answer correct? Is Juve’s?
Find all solutions to the following equations:
- $ = -1 $ b) $ = 1/ $ c) $ = 1 $ d) $ (2) = $ e) $ (7x) = -1 $ f) $ = $ g) $ 2 - 2x = 3 $ h) $ + = 2 $ i) $ (2) + = 0 $
- Find all solutions in the interval $ [0, 4] $ to the following equations:
$ (3x + ) = 1 $ b) $ = $ c) $ + = 1 $ [Hint: Square both sides.*]
$ ^{2}(3x)-5(3x)+4=0 $ [Hint: Think about Example 3.]
- Justify each step in the following derivation of the exact value of $ ^{} $ .
Claim. $ (18^{})=. $
Proof. On the one hand, \(\sin(72^{\circ}) = 2 \sin(36^{\circ}) \cos(36^{\circ}) = 4 \sin(18^{\circ}) \cos(18^{\circ}) [\cos^{2}(18^{\circ}) - \sin^{2}(18^{\circ})] = 4 \sin(18^{\circ}) \cos(18^{\circ}) [1 - 2 \sin^{2}(18^{\circ})].\)
On the other, $ (72{})=(18{}) $ .
Equating the two expressions we’ve found for sin(72°) and then simplifying yields
\[ 1=4\sin(18^{\circ})\left[1-2\sin^{2}(18^{\circ})\right]. \]
Thus, \(\sin(18^{\circ})\) is a number that satisfies the cubic equation \(1 = 4x(1 - 2x^{2})\).
This cubic is equivalent to $ (1 - 2x)(4x^{2} + 2x - 1) = 0 $ , an equation whose solutions are
$ x = $ , $ $ , and $ $ , one of which must be equal to $ (18^{}) $ .
Since $ (18^{}) $ can’t be the first or last of these, we conclude that $ (18^{})= $ .
Draw a right triangle with an $ 18^{} $ angle whose hypotenuse is 4. Use it and the result of Part A to find the values of cosine and tangent at $ 18^{} $ . Then find their values at $ 72^{} $ .
Use the half-angle formula you found in exercise 19 to find an exact expression for $ ^{} $ .
Use the fact that 6 = 15 - 9 and the subtraction identity for sine to find $ ^{} $ exactly.
Find $ ^{} $ exactly. $ ^{} $
In exercise 39 of Chapter 9, you found $ ^{} $ geometrically. Compare the two arguments.