<meta http-equiv="refresh" content="1; url=/nojavascript/">
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Trigonometry - Second Edition Go to the latest version.

Chapter 3: Trigonometric Identities and Equations

Difficulty Level: At Grade Created by: CK-12

Chapter Outline

Chapter Summary

Chapter Summary

Here are the identities studied in this chapter:

Quotient & Reciprocal Identities

\begin{align*}\tan \theta &= \frac{\sin \theta}{\cos \theta} \qquad \cot \theta = \frac{\cos \theta}{\sin \theta}\\ \csc \theta &= \frac{1}{\sin \theta} \ \sec \theta = \frac{1}{\cos \theta} \ \cot \theta = \frac{1}{\tan \theta} \end{align*}

Pythagorean Identities

@$$\begin{align*}\sin^2 \theta + \cos^2 \theta = 1 && 1 + \cot^2 \theta = \csc^2 \theta && \tan^2 \theta + 1 = \sec^2 \theta \end{align*}@$$

Even & Odd Identities

@$$\begin{align*}\sin (-x) &= -\sin x && \cos (-x) = \cos x && \tan (-x) = -\tan x\\ \csc (-x) & = -\csc x && \sec (-x) = \sec x && \cot (-x) = -\cot x\end{align*}@$$

Co-Function Identities

@$$\begin{align*}& \sin \left ( \frac{\pi}{2} - \theta \right ) = \cos \theta && \cos \left ( \frac{\pi}{2} - \theta \right ) = \sin \theta && \tan \left ( \frac{\pi}{2} - \theta \right ) = \cot \theta\\ & \csc \left ( \frac{\pi}{2} - \theta \right )= \sec \theta && \sec \left ( \frac{\pi}{2} - \theta \right ) = \csc \theta && \cot \left ( \frac{\pi}{2} - \theta \right ) = \tan \theta\end{align*}@$$

Sum and Difference Identities

@$$\begin{align*}\cos(\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta && \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \sin(\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta && \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \tan (\alpha + \beta) & = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} && \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\end{align*}@$$

Double Angle Identities

@$$\begin{align*}\cos(2 \alpha) & = \cos^2 \alpha - \sin^2 \alpha = 2 \cos^2 \alpha - 1 = 1 - 2 \sin^2 \alpha \\ \sin(2 \alpha) & = 2 \sin \alpha \cos \beta \\ \tan (2 \alpha) & = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}\end{align*}@$$

Half Angle Identities

@$$\begin{align*}\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}} && \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}} && \tan \frac{\alpha }{2} = \frac{1 - \cos \alpha}{\sin \alpha} = \frac{\sin \alpha}{1 + \cos \alpha}\end{align*}@$$

Product to Sum & Sum to Product Identities

@$$\begin{align*}\sin a + \sin b & = 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2} && \sin a \sin b = \frac{1}{2} [\cos (a-b) - \cos (a+b)] \\ \sin a - \sin b& = 2 \sin \frac{a-b}{2} \cos \frac{a+b}{2} && \cos a \cos b = \frac{1}{2} [\cos (a-b) + \cos (a+b)] \\ \cos a + \cos b & = 2 \cos \frac{a+b}{2} \cos \frac{a-b}{2} && \sin a \cos b = \frac{1}{2} [\sin (a+b) + \sin (a-b)] \\ \cos a - \cos b & = 2 - 2 \sin \frac{a+b}{2} \sin \frac{a-b}{2} && \cos a \sin b = \frac{1}{2} [\sin (a+b) - \sin (a-b)] \end{align*}@$$

Linear Combination Formula

@$\begin{align*}A \cos x + B \sin x = C \cos (x - D)\end{align*}@$, where @$\begin{align*}C = \sqrt{A^2 + B^2}, \cos D = \frac{A}{C}\end{align*}@$ and @$\begin{align*}\sin D = \frac{B}{C}\end{align*}@$

Review Questions

  1. Find the sine, cosine, and tangent of an angle with terminal side on @$\begin{align*}(-8, 15)\end{align*}@$.
  2. If @$\begin{align*}\sin a = \frac{\sqrt{5}}{3}\end{align*}@$ and @$\begin{align*}\tan a < 0\end{align*}@$, find @$\begin{align*}\sec a\end{align*}@$.
  3. Simplify: @$\begin{align*}\frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x}\end{align*}@$.
  4. Verify the identity: @$\begin{align*}\frac{1 + \sin x}{\cos x \sin x} = \sec x (\csc x + 1)\end{align*}@$

For problems 5-8, find all the solutions in the interval @$\begin{align*}[0, 2\pi)\end{align*}@$.

  1. @$\begin{align*}\sec \left (x + \frac{\pi}{2} \right ) + 2 = 0\end{align*}@$
  2. @$\begin{align*}8 \sin \left (\frac{x}{2} \right ) - 8 = 0\end{align*}@$
  3. @$\begin{align*}2 \sin^2 x + \sin 2x =0\end{align*}@$
  4. @$\begin{align*}3 \tan^2 2x = 1\end{align*}@$
  5. Solve the trigonometric equation @$\begin{align*}1 - \sin x = \sqrt{3} \sin x\end{align*}@$ over the interval @$\begin{align*}[0, \pi]\end{align*}@$.
  6. Solve the trigonometric equation @$\begin{align*}2 \cos 3x - 1 = 0\end{align*}@$ over the interval @$\begin{align*}[0, 2\pi]\end{align*}@$.
  7. Solve the trigonometric equation @$\begin{align*}2 \sec^2 x - \tan^4 x = 3\end{align*}@$ for all real values of @$\begin{align*}x\end{align*}@$.

Find the exact value of:

  1. @$\begin{align*}\cos 157.5^\circ\end{align*}@$
  2. @$\begin{align*}\sin \frac{13 \pi}{12}\end{align*}@$
  3. Write as a product: @$\begin{align*}4(\cos 5x + \cos 9x)\end{align*}@$
  4. Simplify: @$\begin{align*}\cos(x - y) \cos y - \sin(x - y) \sin y\end{align*}@$
  5. Simplify: @$\begin{align*}\sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right )\end{align*}@$
  6. Derive a formula for @$\begin{align*}\sin 6x\end{align*}@$.
  7. If you solve @$\begin{align*}\cos 2x = 2 \cos^2x - 1\end{align*}@$ for @$\begin{align*}\cos^2 x\end{align*}@$, you would get @$\begin{align*}\cos^2 x = \frac{1}{2} (\cos 2x + 1)\end{align*}@$. This new formula is used to reduce powers of cosine by substituting in the right part of the equation for @$\begin{align*}\cos^2 x\end{align*}@$. Try writing @$\begin{align*}\cos^4 x\end{align*}@$ in terms of the first power of cosine.
  8. If you solve @$\begin{align*}\cos 2x = 1 - 2 \sin^2 x\end{align*}@$ for @$\begin{align*}\sin^2x\end{align*}@$, you would get @$\begin{align*}\sin^2 x = \frac{1}{2} (1 - \cos 2x)\end{align*}@$. Similar to the new formula above, this one is used to reduce powers of sine. Try writing @$\begin{align*}\sin^4x\end{align*}@$ in terms of the first power of cosine.
  9. Rewrite in terms of the first power of cosine:
    1. @$\begin{align*}\sin^2x \cos^2 2x\end{align*}@$
    2. @$\begin{align*}\tan^4 2x\end{align*}@$

Review Answers

  1. If the terminal side is on @$\begin{align*}(-8,15)\end{align*}@$, then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, @$\begin{align*}c = \sqrt{(-8)^2 + 15^2}\end{align*}@$). Therefore, @$\begin{align*}\sin x = \frac{15}{17}, \cos x = - \frac{8}{17}\end{align*}@$, and @$\begin{align*}\tan x = - \frac{15}{8}\end{align*}@$.
  2. If @$\begin{align*}\sin a = \frac{\sqrt{5}}{3}\end{align*}@$ and @$\begin{align*}\tan a < 0\end{align*}@$, then @$\begin{align*}a\end{align*}@$ is in Quadrant II. Therefore @$\begin{align*}\sec a\end{align*}@$ is negative. To find the third side, we need to do the Pythagorean Theorem. @$$\begin{align*}\left (\sqrt{5} \right )^2 + b^2 & = 3^2 \\ 5 + b^2 & = 9\\ b^2 & = 4 \\ b & = 2\end{align*}@$$ So @$\begin{align*}\sec a = -\frac{3}{2}\end{align*}@$.
  3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn't hold true for values of @$\begin{align*}x\end{align*}@$ that are @$\begin{align*}\frac{\pi}{4} + \frac{n\pi}{2}\end{align*}@$, where @$\begin{align*}n\end{align*}@$ is a positive integer,, since the original expression is undefined for these values of @$\begin{align*}x\end{align*}@$. @$$\begin{align*}& \qquad \qquad \frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x} \\ & \frac{(\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x)}{\cos^2 x - \sin^2 x} \\ & \qquad \qquad \frac{\cos^2 x + \sin^2 x}{1} \\ & \qquad \qquad \frac{1}{1} \\ & \qquad \qquad 1\end{align*}@$$
  4. Change secant and cosecant into terms of sine and cosine, then find a common denominator. @$$\begin{align*}\frac{1 + \sin x}{\cos x \sin x} & = \sec x (\csc x + 1) \\ & = \frac{1}{\cos x} \left (\frac{1}{\sin x} + 1 \right ) \\ & = \frac{1}{\cos x} \left (\frac{1 + \sin x}{\sin x} \right ) \\ & = \frac{1 + \sin x}{\cos x \sin x}\end{align*}@$$
  5. @$$\begin{align*}\sec \left (x + \frac{\pi}{2} \right ) + 2 & = 0 \\ \sec \left (x + \frac{\pi}{2} \right ) & = -2 \\ \cos \left (x + \frac{\pi}{2} \right ) & = - \frac{1}{2} \\ x + \frac{\pi}{2} & = \frac{2 \pi}{3}, \frac{4 \pi}{3} \\ x & = \frac{2 \pi}{3} - \frac{\pi}{2}, \frac{4 \pi}{3} - \frac{\pi}{2} \\ x & = \frac{\pi}{6}, \frac{5 \pi}{6}\end{align*}@$$
  6. @$$\begin{align*}8 \sin \left (\frac{x}{2} \right ) - 8 & = 0 \\ 8 \sin \frac{x}{2} & = 8 \\ \sin \frac{x}{2} & = 1 \\ \frac{x}{2} & = \frac{x}{2} \\ x & = \pi\end{align*}@$$
  7. @$$\begin{align*}& \qquad \ \ 2 \sin^2 x + \sin 2x = 0 \\ & 2 \sin^2 x + 2 \sin x \cos x = 0 \\ & \ 2 \sin x (\sin x + \cos x) = 0 \\ & \ \text{So},\ 2 \sin x = 0 \qquad \quad \text{or}\qquad \quad \sin x + \cos x = 0 \\ & \qquad 2 \sin x = 0 \qquad \qquad \qquad \quad \sin x + \cos x = 0 \\ & \qquad \ \ \sin x = 0 \qquad \qquad \qquad \qquad \qquad \ \sin x = - \cos x \\ & \qquad \qquad \ x = 0, \pi \qquad \qquad \qquad \qquad \qquad \quad x = \frac{3\pi}{4}, \frac{7 \pi}{4}\end{align*}@$$
  8. @$$\begin{align*}3\tan^2 2x & = 1 \\ \tan^2 2x & = \frac{1}{3} \\ \tan 2x & = \pm\frac{\sqrt{3}}{3} \\ 2x & = \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}, \frac{19 \pi}{6}, \frac{23 \pi}{6} \\ x & = \frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}, \frac{19 \pi}{12}, \frac{23 \pi}{12}\end{align*}@$$
  9. @$$\begin{align*}1 - \sin x & = \sqrt{3} \sin x \\ 1 & = \sin x + \sqrt{3} \sin x \\ 1 & = \sin x \left (1 + \sqrt{3} \right ) \\ \frac{1}{1 + \sqrt{3}} & = \sin x \end{align*}@$$ @$\begin{align*}\sin^{-1} \left (\frac{1}{1 + \sqrt{3}} \right ) = x\end{align*}@$ or @$\begin{align*}x = .3747\end{align*}@$ radians and @$\begin{align*}x = 2.7669\end{align*}@$ radians
  10. Because this is @$\begin{align*}\cos 3x\end{align*}@$, you will need to divide by 3 at the very end to get the final answer. This is why we went beyond the limit of @$\begin{align*}2\pi\end{align*}@$ when finding @$\begin{align*}3x\end{align*}@$. @$$\begin{align*}2 \cos 3x - 1 & = 0 \\ 2 \cos 3x & = 1 \\ \cos 3x & = \frac{1}{2} \\ 3x & = \cos^{-1} \left (\frac{1}{2} \right ) = \frac{\pi}{3},\frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{11 \pi}{3}, \frac{13 \pi}{3}, \frac{17 \pi}{3} \\ x & = \frac{\pi}{9},\frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \end{align*}@$$
  11. Rewrite the equation in terms of tan by using the Pythagorean identity, @$\begin{align*}1 + \tan^2 \theta = \sec^2 \theta\end{align*}@$. @$$\begin{align*}2 \sec^2 x - \tan^4 x & = 3 \\ 2 (1 + \tan^2 x) - \tan^4 x & = 3 \\ 2 + 2 \tan^2 x - \tan^4 x & = 3 \\ \tan^4 x - 2 \tan^2 x + 1 & = 0 \\ (\tan^2 x - 1)(\tan^2 x - 1) & = 0\end{align*}@$$ Because these factors are the same, we only need to solve one for @$\begin{align*}x\end{align*}@$. @$$\begin{align*}\tan^2 x - 1 & = 0 \\ \tan^2 x & = 1 \\ \tan x & = \pm 1 \\ x & = \frac{\pi}{4} + \pi k \ \text{and}\ \frac{3 \pi}{4} + \pi k\end{align*}@$$ Where @$\begin{align*}k\end{align*}@$ is any integer.
  12. Use the half angle formula with @$\begin{align*}315^\circ\end{align*}@$. @$$\begin{align*}\cos 157.5^\circ & = \cos \frac{315^\circ}{2} \\ & = - \sqrt{\frac{1 + \cos 315^\circ}{2}} \\ & = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\ & = - \sqrt{\frac{2 + \sqrt{2}}{4}} \\ & = - \frac{\sqrt{2 + \sqrt{2}}}{2} \end{align*}@$$
  13. Use the sine sum formula. @$$\begin{align*}\sin \frac{13 \pi}{12} & = \sin \left (\frac{10 \pi}{12} + \frac{3 \pi}{12} \right ) \\ & = \sin \left (\frac{5 \pi}{6} + \frac{\pi}{4} \right ) \\ & = \sin \frac{5 \pi}{6} \cos \frac{\pi}{4} + \cos \frac{5 \pi}{6} \sin \frac{\pi}{4} \\ & = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + (-\frac{\sqrt{3}}{2}) \cdot \frac{\sqrt{2}}{2} \\ & = \frac{\sqrt{2} - \sqrt{6}}{4}\end{align*}@$$
  14. @$$\begin{align*}4 (\cos 5x + \cos 9x) & = 4 \left [2 \cos \left (\frac{5x + 9x}{2} \right ) \cos \left (\frac{5x - 9x}{2} \right ) \right ] \\ & = 8 \cos 7x \cos (-2x) \\ & = 8 \cos 7x \cos 2x\end{align*}@$$
  15. @$$\begin{align*}& \qquad \qquad \qquad \cos(x - y) \cos y - \sin(x - y) \sin y \\ & \cos y (\cos x \cos y + \sin x \sin y) - \sin y\ (\sin x \cos y - \cos x \sin y) \\ & \cos x \cos^2 y + \sin x \sin y \cos y - \sin x \sin y \cos y + \cos x \sin^2 y \\ & \qquad \qquad \qquad \quad \cos x \cos^2y + \cos x \sin^2 y \\ & \qquad \qquad \qquad \quad \cos x (\cos^2y + \sin^2 y) \\ & \qquad \qquad \qquad \qquad \qquad \cos x\end{align*}@$$
  16. Use the sine and cosine sum formulas. @$$\begin{align*}& \qquad \qquad \qquad \sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right ) \\ & \sin \frac{4 \pi}{3} \cos x - \cos \frac{4 \pi}{3} \sin x + \cos x \cos \frac{5 \pi}{6} - \sin x \sin \frac{5 \pi}{6} \\ & \qquad - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \\ & \qquad \qquad \qquad \qquad \qquad - \sqrt{3} \cos x\end{align*}@$$
  17. Use the sine sum formula as well as the double angle formula. @$$\begin{align*} \sin 6x & = \sin (4x + 2x) \\ & = \sin 4x \cos 2x + \cos 4x \sin 2x \\ & = \sin(2x + 2x) \cos 2x + \cos (2x + 2x) \sin 2x \\ & = \cos 2x\ (\sin 2x \cos 2x + \cos 2x \sin 2x) + \sin 2x(\cos 2x \cos 2x - \sin 2x\ \sin 2x) \\ & = 2 \sin 2x \cos^2 2x + \sin 2x \cos^2 2x - \sin^3 2x \\ & = 3 \sin 2x\ \cos^2 2x - \sin^3 2x \\ & = \sin 2x(3 \cos^2 2x - \sin^2 2x) \\ & = 2 \sin x \cos x [3(\cos^2 x - \sin^2 x)^2 - (2 \sin x \cos x)^2 \\ & = 2 \sin x \cos x [3(\cos^4 x - 2 \sin^2 x \cos^2 x + \sin^4 x) - 4\ \sin^2 x \cos^2 x] \\ & = 2 \sin x \cos x [3 \cos^4 x - 6 \sin^2 x \cos^2 x + 3 \sin^4 x - 4\ \sin^2 x \cos^2 x] \\ & = 2 \sin x \cos x [3 \cos^4 x + 3 \sin^4 x - 10 \sin^2 x \cos^2 x] \\ & = 6 \sin x \cos^5 x + 6 \sin^5 x \cos x - 20 \sin^3x \cos^3x\end{align*}@$$
  18. Using our new formula, @$\begin{align*}\cos^4 x = \left [\frac{1}{2}(\cos 2x + 1) \right ]^2 \end{align*}@$ Now, our final answer needs to be in the first power of cosine, so we need to find a formula for @$\begin{align*}\cos^2 2x\end{align*}@$. For this, we substitute @$\begin{align*}2x\end{align*}@$ everywhere there is an @$\begin{align*}x\end{align*}@$ and the formula translates to @$\begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}@$. Now we can write @$\begin{align*}\cos^4 x\end{align*}@$ in terms of the first power of cosine as follows. @$$\begin{align*}\cos^4 x&=[\frac{1}{2}(\cos 2x+1)]^2 \\ &=\frac{1}{4}(\cos^2 2x+2\cos 2x+1) \\ &=\frac{1}{4}(\frac{1}{2}(\cos 4x+1)+2\cos 2x+1) \\ &=\frac{1}{8}\cos 4x+\frac{1}{8}+\frac{1}{2}\cos 2x+\frac{1}{4} \\ &=\frac{1}{8}\cos 4x+\frac{1}{2}\cos 2x+\frac{3}{8} \end{align*}@$$
  19. Using our new formula, @$\begin{align*}\sin^4 x = \left [\frac{1}{2}(1 - \cos 2x) \right ]^2 \end{align*}@$ Now, our final answer needs to be in the first power of cosine, so we need to find a formula for @$\begin{align*}\cos^2 2x\end{align*}@$. For this, we substitute @$\begin{align*}2x\end{align*}@$ everywhere there is an @$\begin{align*}x\end{align*}@$ and the formula translates to @$\begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}@$. Now we can write @$\begin{align*}\sin^4 x\end{align*}@$ in terms of the first power of cosine as follows. @$$\begin{align*}\sin^4 x&=[\frac{1}{2}(1-\cos 2x)]^2 \\ &=\frac{1}{4}(1-2\cos 2x + \cos^2 2x) \\ &=\frac{1}{4}(1-2\cos 2x + \frac{1}{2}(\cos 4x + 1)) \\ &=\frac{1}{4} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x + \frac{1}{8} \\ &=\frac{1}{8}\cos 4x - \frac{1}{2}\cos 2x + \frac{3}{8} \end{align*}@$$
  20. (a) First, we use both of our new formulas, then simplify: @$$\begin{align*}\sin^2 x \cos^2 2x & = \frac{1}{2}(1 - \cos 2x) \frac{1}{2}(\cos 4x + 1) \\ & = \left (\frac{1}{2} - \frac{1}{2} \cos 2x \right ) \left (\frac{1}{2} \cos 4x + \frac{1}{2} \right ) \\ & = \frac{1}{4} \cos 4x + \frac{1}{4} - \frac{1}{4} \cos 2x \cos 4x - \frac{1}{4} \cos 2x \\ & = \frac{1}{4} (1 - \cos 2x + \cos 4x - \cos 2x \cos 4x) \end{align*}@$$ (b) For tangent, we use the identity @$\begin{align*}\tan x = \frac{\sin x}{\cos x}\end{align*}@$ and then substitute in our new formulas. @$\begin{align*}\tan^4 2x = \frac{\sin^4 2x}{\cos^4 2x} \rightarrow\end{align*}@$ Now, use the formulas we derived in #18 and #19.
  21. @$$\begin{align*}\tan^4 2x &= \frac{\sin^4 2x}{\cos^4 2x} \\ &=\frac{\frac{1}{8}\cos 8x - \frac{1}{2}\cos 4x + \frac{3}{8}}{\frac{1}{8}\cos 8x+\frac{1}{2}\cos 4x+\frac{3}{8}} \\ &=\frac{\cos 8x - 4\cos 4x + 3}{\cos 8x+4\cos 4x+3} \end{align*}@$$

Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9701.

Image Attributions

Files can only be attached to the latest version of None
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 

Original text