Sum, Difference, Multiple, and Half-Angle Formulas
Sum and Difference Formulas
Sum and difference formulas for sines, cosines, and tangents: Let u and v be any real numbers; then
\[ \sin(u+v)=\sin u\cos v+\cos u\sin v \]
\[ \sin(u-v)=\sin u\cos v-\cos u\sin v \]
\[ \cos(u+v)=\cos u\cos v-\sin u\sin v \]
\[ \cos(u-v)=\cos u\cos v+\sin u\sin v \]
\[ \tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v} \]
\[ \tan(u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v} \]
EXAMPLE 24.1 Calculate an exact value for $ $ .
Noting that \(\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}\), apply the difference formula for sines:
\[ \begin{aligned}\sin\frac{\pi}{12}&=\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\&=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}\\&=\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}-\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\\&=\frac{\sqrt{3}-1}{2\sqrt{2}}or\frac{\sqrt{6}-\sqrt{2}}{4}\end{aligned} \]
Cofunction Formulas
Cofunction formulas for the trigonometric functions: Let \(\theta\) be any real number; then
\[ \sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta \]
\[ \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta \]
\[ \tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta \]
\[ \csc\left(\frac{\pi}{2}-\theta\right)=\sec\theta \]
\[ \sec\left(\frac{\pi}{2}-\theta\right)=\csc\theta \]
\[ \cot\left(\frac{\pi}{2}-\theta\right)=\tan\theta \]
Double-Angle Formulas
Double-angle formulas for sines, cosines, and tangents: Let $ $ be any real number; then
\[ \sin2\theta=2\sin\theta\cos\theta\qquad\cos2\theta=\cos^{2}\theta-\sin^{2}\theta\qquad\tan2\theta=\frac{2\tan\theta}{1-\tan^{2}\theta} \]
Also.
\[ \cos2\theta=2\cos^{2}\theta-1=1-2\sin^{2}\theta \]
EXAMPLE 24.2 Given cos θ = $ $ , find cos 2θ.
Use a double-angle formula for cosine: cos2θ = 2cos²θ - 1 = 2(2/3)² - 1 = -1/9
Half-Angle Identities
Half-angle identities for sine and cosine: Let u be any real number; then
\[ \sin^{2}u=\frac{1-\cos2u}{2}\qquad\cos^{2}u=\frac{1+\cos2u}{2} \]
Half-Angle Formulas
Half-angle formulas for sine, cosine, and tangent: Let A be any real number; then
\[ \begin{aligned}\sin\frac{A}{2}=(\pm)\sqrt{\frac{1-\cos A}{2}}\quad&\cos\frac{A}{2}=(\pm)\sqrt{\frac{1+\cos A}{2}}\quad&\tan\frac{A}{2}&=(\pm)\sqrt{\frac{1-\cos A}{1+\cos A}}\\ &=\frac{1-\cos A}{\sin A}\\ &=\frac{\sin A}{1+\cos A}\end{aligned} \]
The sign of the square root in these formulas cannot be specified in general; in any particular case, it is determined by the quadrant in which A/2 lies.
EXAMPLE 24.3 Given $ =,<<2$ , find $ $ and $ $ .
Use the half-angle formulas for sine and cosine. Since $ < < 2$ , dividing all sides of this inequality by 2 yields $ < < $ . Therefore, $ $ lies in quadrant II and the sign of $ $ is to be chosen positive, while the sign of $ $ is to be chosen negative.
\[ \sin\frac{\theta}{2}=\text{}+{\sqrt{\frac{1-\frac{2}{3}}{2}}}=\sqrt{\frac{1}{6}}\qquad\cos\frac{\theta}{2}=\text{}-{\sqrt{\frac{1+\frac{2}{3}}{2}}}={\sqrt{\frac{5}{6}}} \]
Product-To-Sum Formulas
Let u and v be any real numbers:
\[ \sin u\cos v=\frac{1}{2}[\sin\left(u+\nu\right)+\sin\left(u-\nu\right)] \]
\[ \cos u\sin v=\frac{1}{2}[\sin\left(u+v\right)-\sin\left(u-v\right)] \]
\[ \cos u\cos v=\frac{1}{2}[\cos\left(u+v\right)+\cos\left(u-v\right)] \]
\[ \sin u\sin v=\frac{1}{2}[\cos\left(u-v\right)-\cos\left(u+v\right)] \]
Sum-To-Product Formulas
Let a and b be any real numbers:
\[ \sin a+\sin b=2\sin\frac{a+b}{2}\cos\frac{a-b}{2} \]
\[ \cos a+\cos b=2\cos\frac{a+b}{2}\cos\frac{a-b}{2} \]
\[ \sin a-\sin b=2\cos\frac{a+b}{2}\sin\frac{a-b}{2} \]
\[ \cos a-\cos b=-2\sin\frac{a+b}{2}\sin\frac{a-b}{2} \]
EXAMPLE 24.4 Express sin 10x - sin 6x as a product.
Use the formula for $ a - b $ with a = 10x and b = 6x.
\[ \sin10x-\sin6x=2\cos\frac{10x+6x}{2}\sin\frac{10x-6x}{2}=2\cos8x\sin2x \]
SOLVED PROBLEMS
24.1. Derive the difference formula for cosines.
Let u and v be any two real numbers. Shown in Fig. 24-1 is a case for which u, v, and u - v are positive.
The arc with endpoints $ P(u) $ and $ P(v) $ , shown on the left-hand unit circle, has length u - v. The arc with endpoints $ (1, 0) $ and $ P(u - v) $ , shown on the right-hand unit circle, has the same length. Since congruent arcs on congruent circles have congruent chords, the distance from $ P(v) $ to $ P(u) $ must equal the distance from $ (1, 0) $ to $ P(u - v) $ . Hence, from the distance formula,
\[ \sqrt{(\cos u-\cos v)^{2}+(\sin u-\sin v)^{2}}=\sqrt{(\cos(u-v)-1)^{2}+(\sin(u-v)-0)^{2}} \]
Squaring both sides and expanding the squared expressions yields:
\[ \cos^{2}u-2\cos u\cos v+\cos^{2}v+\sin^{2}u-2\sin u\sin v+\sin^{2}v=\cos^{2}(u-v)-2\cos(u-v)+1+\sin^{2}(u-v) \]
Since $ ^{2}u + ^{2}u = 1 $ , $ ^{2}v + ^{2}v = 1 $ , and $ ^{2}(u - v) + ^{2}(u - v) = 1 $ (applying the Pythagorean identity three times), this simplifies to:
\[ 2-2\cos u\cos v-2\sin u\sin v=2-2\cos(u-v) \]
Subtracting 2 from each side and dividing by -2 yields:
\[ \cos u\cos v+\sin u\sin v=\cos(u-v) \]
as required. The proof can be extended to cover all cases, u and v any real numbers.
24.2. Derive
the sum formula for cosines; (b) the cofunction formulas for sines and cosines.
Start with the difference formula for cosines and replace v with -v.
\[ \cos(u-v)=\cos u\cos v+\sin u\sin v \]
\[ \cos[u-(-v)]=\cos u\cos(-v)+\sin u\sin(-v) \]
Now apply the identities for negatives, $ (-v) = v $ and $ (-v) = -v $ , and simplify:
\[ \cos(u+v)=\cos u\cos v+\sin u(-\sin v) \]
\[ \cos(u+v)=\cos u\cos v-\sin u\sin v \]
- Use the difference formula for cosines, with $ u = $ and $ v = $ :
\[ \cos\left(\frac{\pi}{2}-\theta\right)=\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta \]
Since $ (/2)=0 $ and $ (/2)=1 $ , it follows that
\[ \cos\left(\frac{\pi}{2}-\theta\right)=0\cos\theta+1\sin\theta=\sin\theta. \]
Now replace \(\theta\) with \(\pi/2 - \theta\):
\[ \cos\left[\frac{\pi}{2}-\left(\frac{\pi}{2}-\theta\right)\right]=\sin\left(\frac{\pi}{2}-\theta\right) \]
Simplifying yields
\[ \cos\theta=\sin\left(\frac{\pi}{2}-\theta\right) \]
24.3. Derive the difference formula for sines
Start with a cofunction formula, for example, sin θ = cos(π/2 − θ) and replace θ with u − v:
\[ \sin(u-v)=\cos\Big[\frac{\pi}{2}-(u-v)\Big]=\cos\Big(\frac{\pi}{2}-u+v\Big)=\cos\Big[\Big(\frac{\pi}{2}-u\Big)+v\Big] \]
In the last expression, apply the sum formula for cosines with u replaced by $ /2 - u $ .
\[ \sin(u-v)=\cos\left[\left(\frac{\pi}{2}-u\right)+v\right]=\cos\left(\frac{\pi}{2}-u\right)\cos v-\sin\left(\frac{\pi}{2}-u\right)\sin v=\sin u\cos v-\cos u\sin v \]
using the cofunction formulas again at the last step.
24.4. Derive the difference formula for tangents
Start with the quotient identity for $ (u - v) $ .
\[ \tan(u-v)=\frac{\sin\left(u-v\right)}{\cos\left(u-v\right)}=\frac{\sin u\cos v-\cos u\sin v}{\cos u\cos v+\sin u\sin v} \]
To obtain the required expression in terms of tangents, divide numerator and denominator of the last expression by the quantity $ u v $ and apply the quotient identity again:
\[ \tan(u-v)=\frac{\sin u\cos v-\cos u\sin v}{\cos u\cos v+\sin u\sin v}=\frac{\frac{\sin u\cos v}{\cos u\cos v}-\frac{\cos u\sin v}{\cos u\cos v}}{\frac{\cos u\cos v}{\cos u\cos v}+\frac{\sin u\sin v}{\cos u\cos v}}=\frac{\frac{\sin u}{\cos u}-\frac{\sin v}{\cos v}}{1+\frac{\sin u}{\cos u}\frac{\sin v}{\cos v}}=\frac{\tan u-\tan v}{1+\tan u\tan v} \]
24.5. Given $ x = $ , x in quadrant I, and $ y = $ , y in quadrant IV, find (a) $ (x + y) $ ; (b) $ (x + y) $ ;
the quadrant in which $ x + y $ must lie.
From the sum formula for cosines, $ (x + y) = x y - x y $ . $ x $ and $ y $ are given; $ x $ and $ y $ must be determined.
Since $ x = 3/5 $ and x is in quadrant I, $ x = + = 4/5 $ . Since $ y = 2/3 $ and y is in quadrant IV, $ y = - = -()/3 $ . Hence,
\[ \cos(x+y)=\cos x\cos y-\sin x\sin y=\frac{4}{5}\cdot\frac{2}{3}-\frac{3}{5}\bigg(-\frac{\sqrt{5}}{3}\bigg)=\frac{8+3\sqrt{5}}{15} \]
- $ (x + y) $ can be found from the sum formula for tangents, using the given quantities and noting that
\[ \tan x=\frac{\sin x}{\cos x}=\frac{3/5}{4/5}=\frac{3}{4}and\tan y=\frac{\sin y}{\cos y}=\frac{-(\sqrt{5})/3}{2/3}=-\frac{\sqrt{5}}{2}.Hence, \]
\[ \tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{3}{4}+\left(-\frac{\sqrt{5}}{2}\right)}{1-\frac{3}{4}\left(-\frac{\sqrt{5}}{2}\right)}=\frac{6-4\sqrt{5}}{8+3\sqrt{5}} \]
- Since $ (x + y) $ is positive and $ (x + y) $ is negative, $ x + y $ must lie in quadrant IV.
24.6. Derive the double-angle formulas for sine and cosine
Start with the sum formulas for sines and let $ u = v = $ . Then
\[ \sin2\theta=\sin(\theta+\theta)=\sin\theta\cos\theta+\cos\theta\sin\theta=2\sin\theta\cos\theta \]
Similarly for cosines:
\[ \cos2\theta=\cos(\theta+\theta)=\cos\theta\cos\theta-\sin\theta\sin\theta=\cos^{2}\theta-\sin^{2}\theta \]
To derive the other forms of the double-angle formula for cosine, apply Pythagorean identities.
\[ \cos2\theta=\cos^{2}\theta-\sin^{2}\theta=1-\sin^{2}\theta-\sin^{2}\theta=1-2\sin^{2}\theta \]
Also
\[ \cos2\theta=\cos^{2}\theta-\sin^{2}\theta=\cos^{2}\theta-(1-\cos^{2}\theta)=2\cos^{2}\theta-1 \]
24.7. Given $ t = $ , t in quadrant III, find $ 2t $ and $ 2t $ .
From the given information, it follows that $ t = - = - = - $ . Hence,
$ t = - $ and $ t = - $ . Now apply the double-angle formulas for sine and cosine:
\[ \sin2t=2\sin t\cos t=2\Big(-\frac{7}{25}\Big)\Big(-\frac{24}{25}\Big)=\frac{336}{625} \]
\[ \cos2t=\cos^{2}t-\sin^{2}t=\Big(-\frac{24}{25}\Big)^{2}-\Big(-\frac{7}{25}\Big)^{2}=\frac{527}{625} \]
To check this, note that $ ^{2}2t + ^{2}2t = ()^{2} + ()^{2} = 1 $ as expected.
24.8. Derive the half-angle identities
Start with the double-angle formulas for cosine, and solve for $ ^{2}$ and $ ^{2}$ . Then
\[ \begin{aligned}2\cos^{2}\theta-1&=\cos2\theta\\2\cos^{2}\theta&=1+\cos2\theta\\\cos^{2}\theta&=\frac{1+\cos2\theta}{2}\end{aligned} \]
\[ \begin{aligned}1-2\sin^{2}\theta&=\cos2\theta\\-2\sin^{2}\theta&=\cos2\theta-1\end{aligned} \]
\[ 2\sin^{2}\theta=1-\cos2\theta \]
\[ \sin^{2}\theta=\frac{1-\cos2\theta}{2} \]
24.9. Use a double-angle identity to derive an expression for $ t $ in terms of cost.
\[ \begin{aligned}\cos4t=\cos2(2t)=2\cos^{2}2t-1&=2(2\cos^{2}t-1)^{2}-1\\&=2(4\cos^{4}t-4\cos^{2}t+1)-1\\&=8\cos^{4}t-8\cos^{2}t+1\end{aligned} \]
24.10. Use a half-angle identity to derive an expression for $ ^{4}t $ in terms of cosines with exponent 1.
Applying a half-angle identity once yields:
\[ \cos^{4}t=(\cos^{2}t)^{2}=\left(\frac{1+\cos2t}{2}\right)^{2}=\frac{1+2\cos2t+\cos^{2}2t}{4} \]
Applying the identity again, this time to $ ^{2}2t $ , yields:
\[ \begin{aligned}\cos^{4}t=\frac{1+2\cos2t+\cos^{2}2t}{4}&=\frac{1+2\cos2t+\frac{1+\cos2(2t)}{2}}{4}\\&=\frac{2+4\cos2t+1+\cos4t}{8}\\&=\frac{3+4\cos2t+\cos4t}{8}\end{aligned} \]
24.11. Derive the half-angle formulas
In each case, start with the half-angle identities and set $ A = 2$ , thus, $ = $ . Then
\[ \begin{aligned}\cos^{2}\frac{A}{2}&=\frac{1+\cos A}{2}\quad&\sin^{2}\frac{A}{2}&=\frac{1-\cos A}{2}\\\cos\frac{A}{2}&=\pm\sqrt{\frac{1+\cos A}{2}}\quad&\sin\frac{A}{2}&=\pm\sqrt{\frac{1-\cos A}{2}}\end{aligned} \]
For tangent, the derivation is more complicated. First derive the first form of the formula by starting with the quotient identity:
\[ \tan\frac{A}{2}=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\frac{\pm\sqrt{\frac{1-\cos A}{2}}}{\pm\sqrt{\frac{1+\cos A}{2}}}=\pm\sqrt{\frac{1-\cos A}{1+\cos A}} \]
Note that for all three formulas the sign cannot be determined in general, but depends on the quadrant in which A/2 lies. To derive the second form of the half-angle formula for tangents and eliminate the sign ambiguity in this single case, eliminate the fraction under the radical symbol:
\[ \begin{aligned}\tan\frac{A}{2}&=\pm\sqrt{\frac{1-\cos A}{1+\cos A}}=\pm\sqrt{\frac{1-\cos A}{1+\cos A}\cdot\frac{1-\cos A}{1-\cos A}}=\pm\sqrt{\frac{(1-\cos A)^{2}}{1-\cos^{2}A}}\\&=\pm\sqrt{\frac{(1-\cos A)^{2}}{\sin^{2}A}}=\pm\left|\frac{1-\cos A}{\sin A}\right|\end{aligned} \]
Thus $ $ and $ $ are always equal in absolute value. To show that in fact these quantities are always equal, it is sufficient to show that they have the same sign for any value of A between 0 and $ 2$ , which can be done as follows: First note that $ 1-A $ is never negative, so the sign of the fractional expression depends only on the sign of $ A $ . If $ 0 A $ , both $ A $ and $ (A/2) $ are nonnegative; if $ A $ , both are nonpositive. Summarizing,
\[ \tan\frac{A}{2}=\frac{1-\cos A}{\sin A} \]
The derivation of the third form is left to the student (Problem 24.26).
24.12. Given $ u = 4 $ , $ < u < $ , find $ $ .
Use the half-angle formula for tangent. Since $ < u < $ , u lies in quadrant III and the signs of $ u $ and $ u $ are to be chosen negative. To find $ u $ and $ u $ , use a Pythagorean identity.
\[ \sec u=-\sqrt{1+\tan^{2}u}=-\sqrt{1+4^{2}}=-\sqrt{17} \]
Hence
\[ \cos u=\frac{1}{\sec u}=-\frac{1}{\sqrt{17}}\qquad and\qquad\sin u=\cos u\tan u=-\frac{1}{\sqrt{17}}\cdot4=-\frac{4}{\sqrt{17}}. \]
From the half-angle formula for tangent,
\[ \tan\frac{u}{2}=\frac{1-\cos u}{\sin u}=\frac{1-(-1/\sqrt{17})}{-4/(\sqrt{17})}=-\frac{\sqrt{17}+1}{4} \]
24.13. Derive the product-to-sum formulas
Start with the sum and difference formulas for sine, and add left sides and right sides:
\[ \begin{aligned}\sin(u+v)&=\sin u\cos v+\cos u\sin v\\\sin(u-v)&=\sin u\cos v-\cos u\sin v\\\sin(u+v)+\frac{}{\sin(u-v)}&=2\sin u\cos v\end{aligned} \]
Dividing both sides by 2 yields
\[ \sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)] \]
as required.
Now start with the difference and sum formulas for cosine, and add left sides and right sides:
\[ \begin{aligned}\cos(u-v)&=\cos u\cos v+\sin u\sin v\\\cos(u+v)&=\cos u\cos v-\sin u\sin v\\\cos\left(u-v\right)+\frac{\cos\left(u+v\right)}{\cos\left(u+v\right)}&=2\cos u\cos v\end{aligned} \]
Dividing both sides by 2 yields
\[ \cos u\cos v=\frac{1}{2}[\cos\left(u-v\right)+\cos\left(u+v\right)] \]
as required.
The other two formulas are obtained similarly, except that the two sides are subtracted instead of being added.
24.14. Use a product-to-sum formula to rewrite $ 5x x $ as a sum.
\[ \begin{aligned}Use\ \cos u\cos v&=\frac{1}{2}[\cos(u-v)+\cos(u+v)]with u=5x and v=x.Then\\&\cos5x\cos x=\frac{1}{2}[\cos(5x-x)+\cos(5x+x)]=\frac{1}{2}(\cos4x+\cos6x)\end{aligned} \]
24.15. Derive the sum-to-product formulas.
Start with the product-to-sum formula \(\sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)]\). Make the substitution
\[ a=u+v and b=u-v.Then a+b=2u and a-b=2v,hence u=\frac{a+b}{2}and v=\frac{a-b}{2}. \]
Thus $ =[a+b] $ . Multiplying by 2 yields $ a+b=2 $ as required.
The other three sum-to-product formulas are derived by performing the same substitutions in the other three product-to-sum formulas.
24.16. Use a sum-to-product formula to rewrite $ 5u + 3u $ as a product.
\[ \begin{aligned}Use\sin a+\sin b&=2\sin\frac{a+b}{2}\cos\frac{a-b}{2}with a=5u and b=3u.Then\\&\sin5u+\sin3u=2\sin\frac{5u+3u}{2}\cos\frac{5u-3u}{2}=2\sin4u\cos u\end{aligned} \]
24.17. Verify the identity sin3θ = 3sinθ − 4sin3θ.
Starting with the left side, use the sum formula for sines to get an expression in terms of $ $ .
\[ \begin{aligned}\sin3\theta&=\sin(\theta+2\theta)\quad&Algebra\\ &=\sin\theta\cos2\theta+\cos\theta\sin2\theta\quad&Sum formula for sines\\ &=\sin\theta(1-2\sin^{2}\theta)+\cos\theta\cdot2\sin\theta\cos\theta\quad&Double-angle formulas\\ &=\sin\theta-2\sin^{3}\theta+2\sin\theta\cos^{2}\theta\quad&Algebra\\ &=\sin\theta-2\sin^{3}\theta+2\sin\theta(1-\sin^{2}\theta)\quad&Pythagorean identity\\ &=\sin\theta-2\sin^{3}\theta+2\sin\theta-2\sin^{3}\theta\quad&Algebra\\ &=3\sin\theta-4\sin^{3}\theta\quad&Algebra\end{aligned} \]
24.18. If $ f(x) = x $ , show that the difference quotient for $ f(x) $ can be written as $ x ( ) + x $ . (See Chapter 9.)
\[ \begin{aligned}\frac{f(x+h)-f(x)}{h}&=\frac{\sin(x+h)-\sin x}{h}& Substitution\\ &=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}&Sum formula for sine\\ &=\frac{\sin x(\cos h-1)+\cos x\sin h}{h}&Algebra\\ &=\sin x\Big(\frac{\cos h-1}{h}\Big)+\cos x\frac{\sin h}{h}&Algebra\\ \end{aligned} \]
24.19. Verify the reduction formulas: (a) $ (+ ) = -$ ; (b) $ (+ ) = -$ .
- Apply the sum formula for sines, then substitute known values:
\[ \sin\left(\theta+\pi\right)=\sin\theta\cos\pi+\cos\theta\sin\pi=\sin\theta(-1)+\cos\theta(0)=-\sin\theta \]
- Proceeding directly as in (a) fails because $ (/2) $ is undefined. However, first applying a quotient identity eliminates this difficulty.
\[ \tan\Big(\theta+\frac{\pi}{2}\Big)=\frac{\sin\Big(\theta+\frac{\pi}{2}\Big)}{\cos\Big(\theta+\frac{\pi}{2}\Big)}=\frac{\sin\theta\cos\frac{\pi}{2}+\cos\theta\sin\frac{\pi}{2}}{\cos\theta\cos\frac{\pi}{2}-\sin\theta\sin\frac{\pi}{2}}=\frac{\sin\theta(0)+\cos\theta(1)}{\cos\theta(0)-\sin\theta(1)}=\frac{\cos\theta}{-\sin\theta}=-\cot\theta \]
24.20. Find all solutions on the interval $ [0, 2) $ for $ t - 2t = 0 $ .
First use the double-angle formula for sines to obtain an equation involving only functions of t, then solve by factoring:
\[ \begin{aligned}\cos t-\sin2t&=0\\\cos t-2\sin t\cos t&=0\\\cos t(1-2\sin t)&=0\\\cos t=0\quad or\quad1-2\sin t&=0\end{aligned} \]
The solutions of $ t = 0 $ on the interval $ [0, 2) $ are $ /2 $ and $ 3/2 $ . The solutions of $ 1 - 2t = 0 $ , that is, $ t = 1/2 $ , on this interval, are $ /6 $ and $ 5/6 $ .
Solutions: $ ,,, $
24.21. Find all solutions on the interval $ [0, 2) $ for $ 5x - 3x = 0 $ .
First use a sum-to-product formula to put the equation into the form ab = 0.
\[ \cos5x-\cos3x=0 \]
\[ -2\sin\left(\frac{5x+3x}{2}\right)\sin\left(\frac{5x-3x}{2}\right)=0 \]
\[ -2\sin4x\sin x=0 \]
\[ \sin4x\sin x=0 \]
\[ \sin4x=0\quad or\quad\sin x=0 \]
The solutions of $ 4x = 0 $ on the interval $ [0, 2) $ are 0, $ /4 $ , $ /2 $ , $ 3/4 $ , $ $ , $ 5/4 $ , $ 3/2 $ , and $ 7/4 $ . The solutions of $ x = 0 $ on this interval are 0 and $ $ , which have already been listed.
SUPPLEMENTARY PROBLEMS
24.22. Derive the sum formulas for sine and tangent.
24.23. Derive the cofunction formulas (a) for tangents and cotangents; (b) for secants and cosecants.
24.24. Use sum or difference formulas to find exact values for (a) $ $ ; (b) $ ^{} $ ; (c) $ (-) $ .
\[ Ans.\quad(a)\frac{1+\sqrt{3}}{2\sqrt{2}}or\frac{\sqrt{2}+\sqrt{6}}{4};(b)\frac{1-\sqrt{3}}{2\sqrt{2}}or\frac{\sqrt{2}-\sqrt{6}}{4};(c)\frac{1-\sqrt{3}}{1+\sqrt{3}}or\sqrt{3}-2 \]
24.25. Given $ u = - $ , u in quadrant III, and $ v = $ , v in quadrant IV, find (a) $ (u + v) $ ; (b) $ (u - v) $ ; (c) $ (v - u) $ .
\[ Ans.\quad(a)\ \frac{-6+7\sqrt{3}}{20};(b)\ \frac{-3\sqrt{21}+2\sqrt{7}}{20};(c)\ \frac{-6-7\sqrt{3}}{3\sqrt{21}-2\sqrt{7}}\ or-\frac{32\sqrt{21}+75\sqrt{7}}{161} \]
24.26. Derive (a) the double-angle formula for tangents; (b) the third form of the half-angle formula for tangents.
24.27. Given $ t = -3, < t < $ , find (a) $ 2t $ ; (b) $ 2t $ ; (c) $ $ ; (d) $ $ .
Ans. (a) $ - $ ; (b) $ $ ; (c) $ $ ; (d) $ $ .
24.28. Use a double-angle identity to derive an expression for (a) $ 4x $ in terms of $ x $ and $ x $ ; (b) $ 6u $ in terms of $ u $ .
\[ Ans.\quad(a)\sin4x=4\sin x\cos x\left(1-2\sin^{2}x\right);(b)\cos6u=32\cos^{6}u-48\cos^{4}u+18\cos^{2}u-1 \]
24.29. Use a half-angle identity to derive an expression in terms of cosines with exponent 1 for (a) $ {2}2t{2}2t $ ; (b) $ ^{4} $ .
\[ Ans.\quad(a)\ \frac{1-\ \cos8t}{8};(b)\ \frac{3-4\cos x+\ \cos2x}{8} \]
24.30. Complete the derivations of the product-to-sum and sum-to-product formulas (see Problems 24.13 and 24.14).
24.31. (a) Write $ t + t $ as a product. (b) Write $ x t $ as a sum.
\[ Ans.\quad(a)2\sin115\pi t\cos5\pi t;(b)\frac{1}{2}\sin\frac{\pi n}{L}(x+kt)+\frac{1}{2}\sin\frac{\pi n}{L}(x-kt) \]
24.32. Verify that the following are identities: (a) $ = x $ ; (b) $ = u - u $ ;
\[ \left(\mathrm{c}\right)1+\tan\alpha\tan\frac{\alpha}{2}=\sec\alpha;(\mathrm{d})\frac{\cos a-\cos b}{\sin a-\sin b}=-\tan\left(\frac{a+b}{2}\right) \]
24.33. Verify the reduction formulas:
- $ (n+ ) = (-1)^{n}$ , for n any integer; (b) $ (n+ ) = (-1)^{n}$ , for n any integer.
24.34. If $ f(x) = x $ , show that the difference quotient for $ f(x) $ can be written as
\[ \cos x\Big(\frac{\cos h-1}{h}\Big)-\sin x\frac{\sin h}{h}. \]
24.35. Find all solutions on the interval $ [0, 2) $ for the following equations:
- $ - = 0 $ ; (b) $ x + 3x = 2x $ .
Ans. (a) $ 0, , , $ ; (b) $ , , , , , $
24.36. Find approximate values for all solutions on the interval $ [0^{}, 360^{}) $ for $ x = 22x $ .
Ans. $ 32.53^{}, 126.38^{}, 233.62^{}, 327.47^{} $