
# 7.E: Trigonometric Identities and Equations (Exercises)


## 7.1: Solving Trigonometric Equations with Identities

In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

### Verbal

1) We know $$g(x)=\cos x$$ is an even function, and $$f(x)=\sin x$$ and $$h(x)=\tan x$$are odd functions. What about $$G(x)=\cos ^2 x$$, $$F(x)=\sin ^2 x$$ and $$H(x)=\tan ^2 x$$? Are they even, odd, or neither? Why?

All three functions, $$F,G,$$ and $$H$$$,$ are even.

This is because

$$F(-x)=\sin(-x)\sin(-x)=(-\sin x)(-\sin x)=\sin^2 x=F(x),G(-x)=\cos(-x)\cos(-x)=\cos x\cos x= cos^2 x=H(-x)=\tan(-x)\tan(-x)=(-\tan x)(-\tan x)=\tan2x=H(x)$$

2) Examine the graph of $$f(x)=\sec x$$ on the interval $$[-\pi ,\pi ]$$$.$How can we tell whether the function is even or odd by only observing the graph of $$f(x)=\sec x$$?

3) After examining the reciprocal identity for $$\sec t$$$,$ explain why the function is undefined at certain points.

When $$\cos t = 0$$$,$ then $$\sec t = 10$$$,$ which is undefined.

4) All of the Pythagorean identities are related. Describe how to manipulate the equations to get from $$\sin^2t+\cos^2t=1$$ to the other forms.

### Algebraic

For the exercises 5-15, use the fundamental identities to fully simplify the expression.

5) $$\sin x \cos x \sec x$$

$$\sin x$$

6) $$\sin(-x)\cos(-x)\csc(-x)$$

7) $$\tan x\sin x+\sec x\cos^2x$$

$$\sec x$$

8) $$\csc x+\cos x\cot(-x)$$

9) $$\dfrac{\cot t+\tan t}{\sec (-t)}$$

$$\csc x$$

10) $$3\sin^3 t\csc t+\cos^2 t+2\cos(-t)\cos t$$

11) $$-\tan(-x)\cot(-x)$$

$$-1$$

12) $$\dfrac{-\sin (-x)\cos x\sec x\csc x\tan x}{\cot x}$$

13) $$\dfrac{1+\tan ^2\theta }{\csc ^2\theta }+\sin ^2\theta +\dfrac{1}{\sec ^\theta }$$

$$\sec^2 x$$

14) $$\left (\dfrac{\tan x}{\csc ^2 x}+\dfrac{\tan x}{\sec ^2 x} \right )\left (\dfrac{1+\tan x}{1+\cot x} \right )-\dfrac{1}{\cos ^2 x}$$

15) $$\dfrac{1-\cos ^2 x}{\tan ^2 x}+2\sin ^2 x$$

$$\sin^2 x+1$$

For the exercises 16-28, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

16) $$\dfrac{\tan x+\cot x}{\csc x}; \cos x$$

17) $$\dfrac{\sec x+\csc x}{1+\tan x}; \sin x$$

$$\dfrac{1}{\sin x}$$

18) $$\dfrac{\cos x}{1+\sin x}+\tan x; \cos x$$

19) $$\dfrac{1}{\sin x\cos x}-\cot x; \cot x$$

$$\dfrac{1}{\cot x}$$

20) $$\dfrac{1}{1-\cos x}-\dfrac{\cos x}{1+\cos x}; \csc x$$

21) $$(\sec x+\csc x)(\sin x+\cos x)-2-\cot x; \tan x$$

$$\tan x$$

22) $$\dfrac{1}{\csc x-\sin x}; \sec x$$ and $$\tan x$$

23) $$\dfrac{1-\sin x}{1+\sin x}-\dfrac{1+\sin x}{1-\sin x}; \sec x$$ and $$\tan x$$

$$-4\sec x \tan x$$

24) $$\tan x; \sec x$$

25) $$\sec x; \cot x$$

$$\pm \sqrt{\dfrac{1}{\cot ^2 x}+1}$$

26) $$\sec x; \sin x$$

27) $$\cot x; \sin x$$

$$\dfrac{\pm \sqrt{1-\sin ^2 x}}{\sin x}$$

28) $$\cot x; \csc x$$

For the exercises 29-33, verify the identity.

29) $$\cos x-\cos^3x=\cos x \sin^2 x$$

\begin{align*} \cos x-\cos^3x &= \cos x (1-\cos^2 x)\\ &= \cos x\sin ^x \end{align*}

30) $$\cos x(\tan x-\sec(-x))=\sin x-1$$

31) $$\dfrac{1+\sin ^2x}{\cos ^2 x}=\dfrac{1}{\cos ^2 x}+\dfrac{\sin ^2x}{\cos ^2 x}=1+2\tan ^2x$$

\begin{align*} \dfrac{1+\sin ^2x}{\cos ^2 x} &= \dfrac{1}{\cos ^2 x}+\dfrac{\sin ^2x}{\cos ^2 x}\\ &= \sec ^2x+\tan ^2x\\ &= \tan ^2x+1+\tan ^2x\\ &= 1+2\tan ^2x \end{align*}

32) $$(\sin x+\cos x)^2=1+2 \sin x\cos x$$

33) $$\cos^2x-\tan^2x=2-\sin^2x-\sec^2x$$

\begin{align*} \cos^2x-\tan^2x &= 1-\sin^2x-\left (\sec^2x -1 \right )\\ &= 1-\sin^2x-\sec^2x +1\\ &= 2-\sin^2x-\sec^2x \end{align*}

### Extensions

For the exercises 34-39, prove or disprove the identity.

34) $$\dfrac{1}{1+\cos x}-\dfrac{1}{1-\cos (-x)}=-2\cot x\csc x$$

35) $$\csc^2x(1+\sin^2x)=\cot^2x$$

False

36) $$\left (\dfrac{\sec ^2(-x)-\tan ^2x}{\tan x} \right )\left (\dfrac{2+2\tan x}{2+2\cot x} \right )-2\sin ^2x=\cos 2x$$

37) $$\dfrac{\tan x}{\sec x}\sin (-x)=\cos ^2x$$

False

38) $$\dfrac{\sec (-x)}{\tan x+\cot x}=-\sin (-x)$$

39) $$\dfrac{1+\sin x}{\cos x}=\dfrac{\cos x}{1+\sin (-x)}$$

Proved with negative and Pythagorean identities

For the exercises 40-, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

40) $$\dfrac{\cos ^2 \theta -\sin ^2 \theta }{1-\tan ^\theta }=\sin ^2 \theta$$

41) $$3\sin^2\theta + 4\cos^2\theta =3+\cos^2\theta$$

True

\begin{align*} 3\sin^2\theta + 4\cos^2\theta &= 3\sin ^2\theta +3\cos ^2\theta +\cos^2\theta \\ &= 3\left ( \sin ^2\theta +\cos ^2\theta \right )+\cos^2\theta \\ &= 3+\cos^2\theta \end{align*}

42) $$\dfrac{\sec \theta +\tan \theta }{\cot \theta+\cos ^\theta }=\sec ^2 \theta$$

## 7.2: Sum and Difference Identities

In this section, we will learn techniques that will enable us to solve useful problems. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity.

### Verbal

1) Explain the basis for the cofunction identities and when they apply.

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures $$x$$$,$ the second angle measures $$\dfrac{\pi }{2}-x$$$.$Then $$\sin x=\cos \left (\dfrac{\pi }{2}-x \right )$$$.$The same holds for the other cofunction identities. The key is that the angles are complementary.

2) Is there only one way to evaluate $$\cos \left (\dfrac{5\pi }{4} \right )$$$?$Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

3) Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for $$f(x)=\sin (x)$$ and $$g(x)=\cos (x)$$$.$(Hint: $$0-x=-x$$)

$$\sin (-x)=-\sin x$$, so $$\sin x$$ is odd. $$\cos (-x)=\cos (0-x)=\cos x$$, so $$\cos x$$ is even.

### Algebraic

For the exercises 4-9, find the exact value.

4) $$\cos \left (\dfrac{7\pi }{12} \right)$$

5) $$\cos \left (\dfrac{\pi }{12} \right)$$

$$\dfrac{\sqrt{2}+\sqrt{6}}{4}$$

6) $$\sin \left (\dfrac{5\pi }{12} \right)$$

7) $$\sin \left (\dfrac{11\pi }{12} \right)$$

$$\dfrac{\sqrt{6}-\sqrt{2}}{4}$$

8) $$\tan \left (-\dfrac{\pi }{12} \right)$$

9) $$\tan \left (\dfrac{19\pi }{12} \right)$$

$$-2-\sqrt{3}$$

For the exercises 10-13, rewrite in terms of $$\sin x$$ and $$\cos x$$

10) $$\sin \left (x+\dfrac{11\pi }{6} \right)$$

11) $$\sin \left (x-\dfrac{3\pi }{4} \right)$$

$$-\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x$$

12) $$\cos \left (x-\dfrac{5\pi }{6} \right)$$

13) $$\cos \left (x+\dfrac{2\pi }{3} \right)$$

$$-\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x$$

For the exercises 14-19, simplify the given expression.

14) $$\csc \left (\dfrac{\pi }{2}-t \right)$$

15) $$\sec \left (\dfrac{\pi }{2}-\theta \right)$$

$$\csc \theta$$

16) $$\cot \left (\dfrac{\pi }{2}-x \right)$$

17) $$\tan \left (\dfrac{\pi }{2}-x \right)$$

$$\cot x$$

18) $$\sin(2x)\cos(5x)-\sin(5x)\cos(2x)$$

19) $$\dfrac{\tan \left (\dfrac{3}{2}x \right)-\tan \left (\dfrac{7}{5}x \right)}{1+\tan \left (\dfrac{3}{2}x \right)\tan \left (\dfrac{7}{5}x \right)}$$

$$\tan \left (\dfrac{x}{10} \right)$$

For the exercises 20-21, find the requested information.

20) Given that $$\sin a=\dfrac{2}{3}$$ and $$\cos b=-\dfrac{1}{4}$$$,$ with $$a$$ and $$b$$ both in the interval $$\left [ \dfrac{\pi }{2}, \pi \right )$$$,$ find $$\sin (a+b)$$ and $$\cos (a-b)$$.

21) Given that $$\sin a=\dfrac{4}{5}$$ and $$\cos b=\dfrac{1}{3}$$, with $$a$$ and $$b$$ both in the interval $$\left [ 0, \dfrac{\pi }{2} \right )$$, find $$\sin (a-b)$$ and $$\cos (a+b)$$.

$$\sin (a-b)=\left ( \dfrac{4}{5} \right )\left ( \dfrac{1}{3} \right )-\left ( \dfrac{3}{5} \right )\left ( \dfrac{2\sqrt{2}}{3} \right )=\dfrac{4-6\sqrt{2}}{15}$$

$$\cos (a+b)=\left ( \dfrac{3}{5} \right )\left ( \dfrac{1}{3} \right )-\left ( \dfrac{4}{5} \right )\left ( \dfrac{2\sqrt{2}}{3} \right )=\dfrac{3-8\sqrt{2}}{15}$$

For the exercises 22-24, find the exact value of each expression.

22) $$\sin \left ( \cos^{-1}\left ( 0 \right )- \cos^{-1}\left ( \dfrac{1}{2} \right )\right )$$

23) $$\cos \left ( \cos^{-1}\left ( \dfrac{\sqrt{2}}{2} \right )+ \sin^{-1}\left ( \dfrac{\sqrt{3}}{2} \right )\right )$$

$$\dfrac{\sqrt{2}-\sqrt{6}}{4}$$

24) $$\tan \left ( \sin^{-1}\left ( \dfrac{1}{2} \right )- \cos^{-1}\left ( \dfrac{1}{2} \right )\right )$$

### Graphical

For the exercises 25-32, simplify the expression, and then graph both expressions as functions to verify the graphs are identical.

25) $$\cos \left ( \dfrac{\pi }{2}-x \right )$$

$$\sin x$$

26) $$\sin (\pi -x)$$

27) $$\tan \left ( \dfrac{\pi }{3}+x \right )$$

$$\cot \left ( \dfrac{\pi }{6}-x \right )$$

28) $$\sin \left ( \dfrac{\pi }{3}+x \right )$$

29) $$\tan \left ( \dfrac{\pi }{4}-x \right )$$

$$\cot \left ( \dfrac{\pi }{4}+x \right )$$

30) $$\cos \left ( \dfrac{7\pi }{6}+x \right )$$

31) $$\sin \left ( \dfrac{\pi }{4}+x \right )$$

$$\dfrac{\sin x}{\sqrt{2}}+\dfrac{\cos x}{\sqrt{2}}$$

32) $$\cos \left ( \dfrac{5\pi }{4}+x \right )$$

For the exercises 33-41, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think $$2x=x+x$$)

33) $$f(x)=\sin(4x)-\sin(3x)\cos x, g(x)=\sin x \cos(3x)$$

They are the same.

34) $$f(x)=\cos(4x)+\sin x \sin(3x), g(x)=-\cos x \cos(3x)$$

35) $$f(x)=\sin(3x)\cos(6x), g(x)=-\sin(3x)\cos(6x)$$

They are different, try $$g(x)=\sin(9x)-\cos(3x)\sin(6x)$$

36) $$f(x)=\sin(4x), g(x)=\sin(5x)\cos x-\cos(5x)\sin x$$

37) $$f(x)=\sin(2x), g(x)=2 \sin x \cos x$$

They are the same.

38) $$f(\theta )=\cos(2\theta ), g(\theta )=\cos^2\theta -\sin^2\theta$$

39) $$f(\theta )=\tan(2\theta ), g(\theta )=\dfrac{\tan \theta }{1+\tan^2\theta }$$

They are different, try $$g(\theta )=\dfrac{2\tan \theta }{1-\tan^2\theta }$$

40) $$f(x)=\sin(3x)\sin x, g(x)=\sin^2(2x)\cos^2x-\cos^2(2x)\sin2x$$

41) $$f(x)=\tan(-x), g(x)=\dfrac{\tan x-\tan(2x)}{1-\tan x \tan(2x)}$$

They are different, try $$g(x)=\dfrac{\tan x-\tan(2x)}{1+\tan x \tan(2x)}$$

### Technology

For the exercises 42-46, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.

42) $$\sin (75^{\circ})$$

43) $$\sin (195^{\circ})$$

$$-\dfrac{\sqrt{3}-1}{2\sqrt{2}}$$, or $$-0.2588$$

44) $$\cos (165^{\circ})$$

45) $$\cos (345^{\circ})$$

$$\dfrac{1+\sqrt{3}}{2\sqrt{2}}$$, or $$-0.9659$$

46) $$\tan (-15^{\circ})$$

### Extensions

For the exercises 47-51, prove the identities provided.

47) $$\tan \left ( x+\dfrac{\pi }{4} \right )=\dfrac{\tan x+1}{1-\tan x}$$

\begin{align*} \tan \left ( x+\dfrac{\pi }{4} \right ) &= \\ \dfrac{\tan x + \tan\left (\tfrac{\pi}{4} \right )}{1-\tan x \tan\left (\tfrac{\pi}{4} \right )} &= \\ \dfrac{\tan x+1}{1-\tan x(1)} &= \dfrac{\tan x+1}{1-\tan x} \end{align*}

48) $$\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\sin a \cos a + \sin b \cos b}{\sin a \cos a - \sin b \cos b}$$

49) $$\dfrac{\cos (a+b)}{\cos a \cos b}=1-\tan a \tan b$$

\begin{align*} \dfrac{\cos (a+b)}{\cos a \cos b} &= \\ \dfrac{\cos a \cos b}{\cos a \cos b}- \dfrac{\sin a \sin b}{\cos a \cos b} &= 1-\tan a \tan b \end{align*}

50) $$\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y$$

51) $$\dfrac{\cos(x+h)-\cos(x)}{h}=\cos x\dfrac{\cos h-1}{h}-\sin x \dfrac{\sin h}{h}$$

\begin{align*} \dfrac{\cos(x+h)-\cos(x)}{h} &= \\ \dfrac{\cos x\cosh - \sin x\sinh -\cos x}{h} &= \\ \dfrac{\cos x(\cosh-1) - \sin x(\sinh-1)}{h} &= \cos x\dfrac{\cos h-1}{h}-\sin x \dfrac{\sin h}{h} \end{align*}

For the exercises 52-, prove or disprove the statements.

52) $$\tan (u+v)=\dfrac{\tan u+\tan v}{1-\tan u \tan v}$$

53) $$\tan (u-v)=\dfrac{\tan u-\tan v}{1+\tan u \tan v}$$

54) $$\dfrac{\tan (x+y)}{1+\tan x \tan x}=\dfrac{\tan x + \tan y}{1-\tan^2 x \tan^2 y}$$
55) If $$\alpha ,\beta$$$,$ and $$\gamma$$ are angles in the same triangle, then prove or disprove