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Algebra and Precalculus A Practical Guide

Cameron Geisler

Section 15.4 Double-Angle Identities

Identities about double angles can be derived from the sum and difference identities.
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\begin{align*} \begin{array}{ccc} \boxed{\sin{(2\theta)} = 2\sin{\theta}\cos{\theta}} \amp \quad \boxed{\begin{array}{rl} \cos{(2\theta)} \amp = \cos^2{\theta} - \sin^2{\theta} \\ \amp = 2\cos^2{\theta} - 1 \\ \amp = 1 - 2\sin^2{\theta} \end{array}} \amp \quad \boxed{\tan{(2\theta)} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}} \end{array} \end{align*}
These are expressions for the trigonometric ratios of double angles (\(\sin{(2\theta)}, \cos{(2\theta)}\text{,}\) and \(\tan{(2\theta)}\)), in terms of their associated “single” angles (\(\sin{\theta}, \cos{\theta}\text{,}\) and \(\tan{\theta}\)).
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Subsection 15.4.1 Finding Exact Value Examples

Exercise Group 15.4.1. Evaluate Exact Expressions.

Evaluate each expression exactly, using a double-angle identity.
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(a)
\(2 \sin{\frac{\pi}{12}} \cos{\frac{\pi}{12}}\)
Answer.
\(\frac{1}{2}\)
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(b)
\(\cos^2{\brac{\frac{\pi}{3}}} - \sin^2{\brac{\frac{\pi}{3}}}\)
Answer.
\(-\frac{1}{2}\)
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(c)
\(\frac{2 \tan \frac{\pi}{6}}{1 - \tan^2 \frac{\pi}{6}}\)
Answer.
\(\sqrt{3}\)
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(d)
\(1 - 2 \sin^2 15^\circ\)
Answer.
\(\frac{\sqrt{3}}{2}\)
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(e)
\(2 \sin \frac{\pi}{6} \cos \frac{\pi}{6}\)
Answer.
\(\frac{\sqrt{3}}{2}\)
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Exercise Group 15.4.2. Expand Expressions.

Expand each expression, using a double-angle identity.
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(a)
\(3\sin{4x}\)
Hint.
\(4x = 2 \cdot 2x\) (\(4x\) is double \(2x\)).
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Answer.
\(6\sin{2x}\cos{2x}\)
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(b)
\(4\tan{6x}\)
Hint.
\(6x = 2 \cdot 3x\) (\(6x\) is double \(3x\)).
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Answer.
\(\frac{8\tan{3x}}{1 - \tan^2{3x}}\)
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(c)
\(\frac{1}{2}\sin{\frac{1}{2}x}\)
Hint.
\(\frac{1}{2}x = 2 \cdot \frac{1}{4}x\text{.}\)
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Answer.
\(\sin{\frac{x}{4}}\cos{\frac{x}{4}}\)
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(d)
\(\frac{1}{4}\tan{\frac{1}{2}x}\)
Answer.
\(\frac{1}{2}\cdot\frac{\tan{\frac{x}{4}}}{1 - \tan^2{\frac{x}{4}}}\)
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Subsection 15.4.2 Exact Values with Variable Angles

Exercise Group 15.4.3. Find Exact Values.

Find the exact value of each trigonometric ratio, from the given information.
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(a)
If \(\sin{\theta} = \frac{5}{13}\) and \(\frac{\pi}{2} \lt \theta \lt \pi\text{.}\) Find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)} = -\frac{120}{169}, \cos{(2\theta)} = \frac{119}{169}, \tan{(2\theta)} = -\frac{120}{119}\text{.}\)
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(b)
If \(\cos{\theta}=\frac{2}{5}\) and \(\theta\) is in quadrant IV, find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)}=-\frac{4\sqrt{21}}{25}, \cos{(2\theta)}=-\frac{17}{25}, \tan{(2\theta)}=\frac{4\sqrt{21}}{17}\text{.}\)
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(c)
If \(\sin{\theta} = \frac{12}{13}\text{,}\) where \(\frac{\pi}{2} \lt \theta \lt \pi\text{,}\) then find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)} = -\frac{120}{169}, \cos{(2\theta)} = -\frac{119}{169}, \tan{(2\theta)} = \frac{120}{119}\text{.}\)
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(d)
Let \(\cos{\theta} = -\frac{7}{25}\) and \(\ang{90} \lt \theta \lt \ang{180}\text{.}\) Find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)} = -\frac{336}{625}, \cos{(2\theta)} = -\frac{527}{625}, \tan{(2\theta)} = \frac{336}{527}\text{.}\)
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(e)
Let \(\theta\) be an angle in quadrant III with \(\cos{\theta} = -\frac{8}{17}\text{.}\) Find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)} = \frac{240}{289}, \cos{(2\theta)} = -\frac{161}{289}, \tan{(2\theta)} = -\frac{240}{161}\text{.}\)
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(f)
If \(\sin{\theta} = -\frac{12}{13}\) and \(\theta\) is in quadrant III, find \(\sin{(2\theta)}, \cos{(2\theta)}\) and \(\tan{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)} = \frac{120}{169}, \cos{(2\theta)} = -\frac{119}{169}, \tan{(2\theta)} = -\frac{120}{119}\text{.}\)
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(g)
If \(\cos{\theta} = -\frac{3}{5}\) and \(\theta\) is in quadrant III, find \(\sin{(2\theta)}\text{.}\)
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Answer.
\(\sin{(2\theta)}=\frac{24}{25}\text{.}\)
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Subsection 15.4.3 Simplify Using Double-Angle Identities

Exercise Group 15.4.4. Simplify Expressions I.

Simplify each expression into a single trigonometric function.
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(a)
\(2 \sin{(4x)} \cos{(4x)}\)
Answer.
\(\sin{(8x)}\)
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(b)
\(10 \cos^2{x} - 5\)
Answer.
\(5 \cos{(2x)}\)
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(c)
\(2 \sin{\brac{\frac{1}{2} x}} \cos{\brac{\frac{1}{2} x}}\)
Answer.
\(\sin{x}\)
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(d)
\(1 - 2 \sin^2{(3x)}\)
Answer.
\(\cos{(6x)}\)
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(e)
\(\frac{\sin{(2x)}}{2 \sin{x}}\)
Answer.
\(\cos{x}\)
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(f)
\(6 \sin{(8x)} \cos{(8x)}\)
Answer.
\(3 \sin{(16x)}\)
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(g)
\(6\sin^2{x}-3\)
Answer.
\(-3\cos{(2x)}\)
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(h)
\(4\cos^2{(5x)}-4\sin^2{(5x)}\)
Answer.
\(4\cos{(10x)}\)
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(i)
\(4\cos^2{(2x)}-2\)
Answer.
\(2\cos{(4x)}\)
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(j)
\(4\sin{(3x)}\cos{(3x)}\)
Answer.
\(2\sin{(6x)}\)
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(k)
\(2\cos^2{(8x)}-1\)
Answer.
\(\cos{(16x)}\)
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(l)
\(8\sin{(5x)}\cos{(5x)}\)
Answer.
\(4\sin{(10x)}\)
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(m)
\(4\sin{\brac{\frac{x}{2}}}\cos{\brac{\frac{x}{2}}}\)
Answer.
\(2\sin{x}\)
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(n)
\(2\sin^2{(2x)}-2\cos^2{(2x)}\)
Answer.
\(-2\cos{(4x)}\)
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Exercise Group 15.4.5. Simplify Expressions II.

Simplify each expression into a single trigonometric function.
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(a)
\(\frac{4\tan{(3x)}}{1 - \tan^2{(3x)}}\)
Answer.
\(2 \tan{(6x)}\)
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(b)
\(\frac{\cos{(2x)} - 1}{2 \sin{x}}\)
Hint.
choose the double angle for cosine which cancels out the \(-1\) in the numerator.
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Answer.
\(-\sin{x}\)
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(c)
\(4 \sin^2{(2x)} - 2\)
Answer.
\(-2 \cos{(4x)}\)
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(d)
\(\frac{6\tan{(4x)}}{1-\tan^2{(4x)}}\)
Answer.
\(3\tan{(8x)}\)
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(e)
\(\frac{\cos{(2x)} + 1}{2 \cos{x}}\)
Hint.
choose the double angle for cosine which cancels out the \(1\) in the numerator.
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Answer.
\(\cos{x}\)
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(f)
\(\frac{8\tan{(4x)}}{1-\tan^2{(4x)}}\)
Answer.
\(4\tan{(8x)}\)
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(g)
\(2\sin{(6x)}\brac{\cos^2{(3x)}-\sin^2{(3x)}}\)
Answer.
\(\sin{(12x)}\)
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(h)
\(\frac{1}{4}\sec{(6x)}\csc{(6x)}\)
Answer.
\(\frac{1}{2\sin{(12x)}}\)\(\frac{1}{2}\csc{(12x)}\)
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(i)
\(4\sin^2{\brac{\frac{x}{2}}}-2\)
Answer.
\(-2\cos{x}\)
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(j)
\(\frac{\sin{(6x)}}{2\sin{(3x)}}\)
Answer.
\(\cos{(3x)}\)
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Exercise Group 15.4.6. Simplify Expressions III.

Simplify each expression into a single trigonometric function.
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(a)
\(\frac{\sin{(2x)}}{\cos{(2x)} + 1}\)
Hint.
choose the double angle for cosine which cancels out the 1 in the denominator.
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Answer.
\(\tan{x}\)
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(b)
\(\frac{\cos^3{x}}{\cos{(2x)} + \sin^2{x}}\)
Answer.
\(\cos{x}\)
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(c)
\(\frac{\sin^3{x}}{\cos{(2x)}-\cos^2{x}}\)
Answer.
\(-\sin{x}\)
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(d)
\(\frac{3-6\sin^2{x}}{2\sin{x}\cos{x}}\)
Answer.
\(3 \cot{(2x)}\)
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(e)
\(\brac{\sin^2{x} - \cos^2{x}}^2 - \sin^2{(2x)}\)
Hint.
use a double-angle identity, do not expand.
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Answer.
\(\cos{(4x)}\)
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(f)
\(\frac{2\tan{(2x)}}{3-3\tan^2{(2x)}}\)
Answer.
\(\frac{1}{3}\tan{(4x)}\)
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(g)
\(\sec{(8x)}\brac{\sin^2{(4x)}-\cos^2{(4x)}}\)
Answer.
\(-1\)
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(h)
\(\frac{1}{2}\cot{(4x)}\brac{1-\tan^2{(4x)}}\)
Answer.
\(\cot{(8x)}\)
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(i)
\(\sin{(4x)}\csc{(2x)}-2\cos{(2x)}\)
Answer.
\(0\)
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(j)
\(\sin{(4x)}-\brac{\sin{(2x)}+\cos{(2x)}}^2\)
Answer.
\(-1\)
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(k)
\(\sin^4{(3x)}-\cos^4{(3x)}\)
Answer.
\(-\cos{(6x)}\)
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(l)
\(\frac{2}{1-\cos{(8x)}}\)
Answer.
\(\csc^2{(4x)}\)
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(m)
\(\frac{4}{\tan{(3x)}-\cot{(3x)}}\)
Answer.
\(-2\tan{(6x)}\)
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Subsection 15.4.4 Equations with Double-Angle Identities

Exercise Group 15.4.7. Solve Equations I.

Solve each equation, by finding all solutions in the interval \([0,2\pi)\text{.}\)
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(a)
\(\sin{(2x)}=\cos{x}\)
Answer.
\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2},\frac{3\pi}{2}\)
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(b)
\(\sin{(2x)}-2\cos^2{x}=0\)
Answer.
\(x=\frac{\pi}{4},\frac{5\pi}{4},\frac{\pi}{2},\frac{3\pi}{2}\)
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(c)
\(\sin{(2x)}=\sqrt{2}\cos{x}\)
Answer.
\(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{\pi}{2},\frac{3\pi}{2}\)
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(d)
\(-2\cos{(2x)}=2\sin{x}\)
Answer.
\(x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}\)
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(e)
\(\sin{(2x)}+\cos{x}=0\)
Answer.
\(x=\frac{\pi}{2},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\)
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(f)
\(2\sin{x}\cos{x}=\cos{(2x)}\)
Answer.
\(x=\frac{\pi}{8},\frac{5\pi}{8},\frac{9\pi}{8},\frac{13\pi}{8}\)
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(g)
\(\sin{x}+\cos{(2x)}=1\)
Answer.
\(x=0,\frac{\pi}{6},\frac{5\pi}{6},\pi\)
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(h)
\(\frac{1}{2}\sin{(2x)}-\cos^2{x}=0\)
Answer.
\(x=\frac{\pi}{4},\frac{\pi}{2},\frac{5\pi}{4},\frac{3\pi}{2}\)
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(i)
\(3\sin{x}+\cos{(2x)}=2\)
Answer.
\(x=\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}\)
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(j)
\(\cos{(2x)}=\sin{x}\)
Answer.
\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\)
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(k)
\(\cos{(2x)}-3\sin{x}=2\)
Answer.
\(x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{3\pi}{2}\)
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(l)
\(\cos{(2x)}-5\cos{x}=2\)
Answer.
\(x=\frac{2\pi}{3},\frac{4\pi}{3}\)
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(m)
\(3\cos{(2x)}+2\sin^2{x}=2\)
Answer.
\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\)
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(n)
\(\tan{(2x)}+\tan{x}=0\)
Answer.
\(x=0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3}\)
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Exercise Group 15.4.8. Solve Equations II.

Solve each equation, by finding all solutions in the interval \([0,2\pi)\text{.}\)
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(a)
\(\cos{(4x)}+2\cos^2{(2x)}=2\)
Answer.
\(x=\frac{\pi}{12},\frac{5\pi}{12},\frac{7\pi}{12},\frac{11\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12},\frac{19\pi}{12},\frac{23\pi}{12}\)
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(b)
\(5\cos{(2x)}-\cos{x}+3=0\)
Answer.
\(x=\frac{\pi}{3},\frac{5\pi}{3},1.98,4.30\)
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(c)
\(\sin{(2x)}=\cot{x}\)
Answer.
\(x=\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}\)
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(d)
\(\tan{x}-\cot{x}=2\)
Answer.
\(x=\frac{3\pi}{8},\frac{7\pi}{8},\frac{11\pi}{8},\frac{15\pi}{8}\)
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(e)
\(4\cos{(2x)}+10\sin{x}-7=0\)
Answer.
\(x=\frac{\pi}{6},\frac{5\pi}{6},0.85,2.29\)
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(f)
\(\csc^2{x}=2\sec{(2x)}\)
Answer.
\(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\)
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(g)
\(10\cos{(2x)}-8\cos{x}+1=0\)
Answer.
\(x=\frac{2\pi}{3},\frac{4\pi}{3},0.45,5.83\)
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(h)
\(4\sin^2{x}=2-\cos^2{(2x)}\)
Answer.
\(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\)
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Subsection 15.4.5 Proving Identities with Double Angles

Exercise Group 15.4.9. Prove Identities I.

Prove each trigonometric identity.
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(a)
\(\frac{\brac{\sin{x} + \cos{x}}^2}{\sin{(2x)}} = \csc{(2x)} + 1\)
Hint.
Pythagorean identity, double-angle identity.
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(b)
\(\frac{\csc{x}}{2\cos{x}} = \csc{(2x)}\)
Hint.
reciprocal identity, double-angle identity.
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(c)
\(\sec^2{x}=\frac{2}{1+\cos{(2x)}}\)
Hint.
double-angle identity \(\cos{(2x)}=2\cos^2{x}-1\text{,}\) rewrite in terms of sine and cosine.
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(d)
\(\cot{x}\csc{(2x)}=\frac{1}{2\sin^2{x}}\)
Hint.
rewrite in terms of sine and cosine, double-angle identity, cancel common factors.
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(e)
\(\frac{1+\cos{(2x)}}{\sin{(2x)}}=\cot x\)
Hint.
double-angle identity, cancel common factors.
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(f)
\(1+\sin{(2x)}=(\sin{x}+\cos{x})^2\)
Hint.
expand, Pythagorean identity, double-angle identity.
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(g)
\(\tan{x}=\frac{1-\cos{(2x)}}{\sin{(2x)}}\)
Hint.
double-angle identity, rewrite in terms of sine and cosine.
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(h)
\(\sin{(2x)}\brac{\tan{x}+\cot{x}}=2\)
Hint.
double-angle identity, rewrite in terms of sine and cosine, distribute, cancel common factors.
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(i)
\(\frac{\sin{(2x)}}{1-\cos{(2x)}} = \cot{x}\)
Hint.
double-angle identity, cancel common factors.
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Exercise Group 15.4.10. Prove Identities II.

Prove each trigonometric identity.
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(a)
\(\frac{\sin{x}+\tan{x}}{1+\cos{x}}=\frac{\sin{(2x)}}{2 \cos^2{x}}\)
Hint.
start with both sides, double-angle identity, clear denominators, cancel.
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(b)
\(\frac{2\tan{x}}{1+\tan^2{x}} = \sin{(2x)}\)
Hint.
convert to sine and cosine, clear denominators, Pythagorean identity, double-angle identity.
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(c)
\(\frac{\sin{(6x)}}{1 + \cos{(6x)}} = \tan{(3x)}\)
Hint.
work with both sides, double-angle identity, cancel common factors.
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(d)
\(\tan{(2x)} = \frac{2}{\cot{x} - \tan{x}}\)
Hint.
work with both sides, rewrite in terms of tangent, simplify complex fraction by clearing denominators.
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(e)
\(\cot{x}-\csc{x}=\frac{\cos{(2x)}-\cos{x}}{\sin{(2x)}+\sin{x}}\)
Hint.
work with both sides, double-angle identity, factor quadratic trinomial, cancel common factors, rewrite in terms of sine and cosine, combine fractions (or split fraction).
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(f)
\(\frac{2\cos{x} + 2\cos^2{x}}{\sin{(2x)}} = \frac{\sin{x}}{1 - \cos{x}}\)
Hint.
start with the LHS, double-angle identity, factor out common factor, cancel common factors, multiply by the conjugate.
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(g)
\(\frac{\sin{(2x)}}{1 + \cos{(2x)}} = \frac{\sec^2{x} - 1}{\tan{x}}\)
Hint.
work with both sides, double-angle identity, cancel common factors, Pythagorean identity (or rewrite in terms of sine and cosine, simplify complex fraction by clearing denominators).
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(h)
\((\csc{x}-\sin{x})\tan{x}=\frac{\sin{(2x)}}{2\sin{x}}\)
Hint.
work with both sides, double-angle identity, rewrite in terms of sine and cosine, distribute, Pythagorean identity.
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Exercise Group 15.4.11. Prove Identities III.

Prove each trigonometric identity.
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(a)
\(\cot{x}-\tan{x}=\frac{4\cos^2{x}-2}{\sin{(2x)}}\)
Hint.
work with both sides, double-angle identity, rewrite in terms of sine and cosine, combine fractions.
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(b)
\(\frac{\sin{(4x)}-\sin{(2x)}}{\cos{(4x)}+\cos{(2x)}}=\tan{x}\)
Hint.
work with both sides, double-angle identity, factor quadratic trinomial, cancel common factors, rewrite in terms of sine and cosine.
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(c)
\(\frac{\cot{x}-\cos{x}}{1-\sin{x}} = \frac{\sin{(2x)}}{1-\cos{(2x)}}\)
Hint.
work with both sides, rewrite in terms of sine and cosine, double-angle identity, simplify complex fraction by clearing denominators, cancel common factors.
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(d)
\(\cot{(2x)}=\frac{\cot^2{x}-1}{2\cot{x}}\)
Hint.
start with the RHS, rewrite in terms of sine and cosine, double-angle identity.
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(e)
\(\frac{\cos{(2x)}}{1-\sin{(2x)}}=\frac{1+\tan{x}}{1-\tan{x}}\)
Hint.
work with both sides, double-angle identity, rewrite in terms of sine and cosine, simplify complex fraction, multiply by the conjugate.
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(f)
\(-\sec{(2x)}=\frac{\tan{x}+\cot{x}}{\tan{x}-\cot{x}}\)
Hint.
work with both sides, rewrite in terms of sine and cosine, simplify complex fraction by clearing denominators, double-angle identity, Pythagorean identity.
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Subsection 15.4.6 Advanced Examples

Checkpoint 15.4.1. Geometric Mean.

Find the geometric mean of \(\sin^2\theta\) and \(\cos^2\theta\text{,}\) in terms of sine only. Note: the geometric mean of two numbers \(a\) and \(b\) is \(\sqrt{ab}\text{.}\)
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Answer.
\(\frac{1}{2} \abs{\sin{(2\theta)}}\text{.}\)
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Checkpoint 15.4.2. Exact Value in Terms of Variables.

Given that \(\cos\theta=\frac{a}{x}\) and \(\theta\) is in quadrant 4, find the exact value of \(\sin 2\theta\) in terms of \(a\) and \(x\text{.}\)
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Hint.
\(\sin 2\theta=2\sin\theta\cos\theta\text{.}\)
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Answer.
\(\sin 2\theta=-\frac{2a\sqrt{x^2-a^2}}{x^2}\text{.}\)
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Checkpoint 15.4.3. Expression for Tangent.

Given that \(\csc\theta=\frac{b}{x}\) and \(\theta\) is in quadrant \(2\text{,}\) find an expression for \(\tan 2\theta\) in terms of \(b\) and \(x\text{.}\)
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Hint.
\(\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}\text{,}\) and simplify the complex fraction.
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Answer.
\(\tan 2\theta=-\frac{2x\sqrt{b^2-x^2}}{b^2-2x^2}\text{.}\)
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Checkpoint 15.4.4. Solving with Tangent.

If \(\sin{x} = -2 \cos{x}\) and \(\frac{\pi}{2} \leq x \leq \pi\text{,}\) find \(\tan{(2x)}\text{.}\)
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Hint.
to find \(\tan{x}\text{,}\) divide both sides of the equation by \(\cos{x}\text{.}\)
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Answer.
\(\frac{4}{3}\text{.}\)
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Subsection 15.4.7 Proofs of Double-Angle Identities

The double-angle identities are derived from the sum and difference identities, by adding the same angle \(\theta\) to itself.
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Proof.

Using the sum and difference identities, for \(\sin{(2\theta)}\text{,}\)
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\begin{align*} \sin{(2\theta)} \amp = \sin{(\theta + \theta)}\\ \amp = \sin{\theta}\cos{\theta} + \cos{\theta}\sin{\theta} \amp\amp \text{by the sine sum identity}\\ \amp = 2\sin{\theta}\cos{\theta} \amp\amp \text{simplify} \end{align*}
For \(\cos{(2\theta)}\text{,}\)
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\begin{align*} \cos{(2\theta)} \amp = \cos{(\theta + \theta)}\\ \amp = \cos{\theta}\cos{\theta} - \sin{\theta}\sin{\theta} \amp\amp \text{by the cosine sum identity}\\ \amp = \cos^2{\theta} - \sin^2{\theta} \amp\amp \text{simplify} \end{align*}
Then, for the 2nd form of the cosine identity, use the Pythagorean identity,
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\begin{align*} \cos{(2\theta)} \amp = \cos^2{\theta} - \sin^2{\theta}\\ \amp = (1 - \sin^2{\theta}) - \sin^2{\theta} \amp\amp \text{by the Pythagorean identity}\\ \amp = 1 - 2\sin^2{\theta} \amp\amp \text{simplify} \end{align*}
And, for the 3rd form,
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\begin{align*} \cos{(2\theta)} \amp = \cos^2{\theta} - (1 - \cos^2{\theta}) \amp\amp \text{by the Pythagorean identity}\\ \amp = 2\cos^2{\theta} - 1 \amp\amp \text{simplify} \end{align*}
Also, for \(\tan{(2\theta)}\text{,}\)
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\begin{align*} \tan{(2\theta)} \amp = \tan{(\theta + \theta)}\\ \amp = \frac{\tan{\theta} + \tan{\theta}}{1 - \tan{\theta}\tan{\theta}} \amp\amp \text{by the tangent sum identity}\\ \amp = \frac{2\tan{\theta}}{1 - \tan^2{\theta}} \amp\amp \text{simplify} \end{align*}
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Subsection 15.4.8 Triple-Angle Identities

We can also find expressions for \(\sin{(3\theta)}\) and \(\cos{(3\theta)}\) in terms of their associated single angles, using the sum and difference identities and the double-angle identities. In other words, triple-angle identities.
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Example 15.4.5. Deriving the Cosine Triple-Angle Identity.

To find \(\cos{(3x)}\text{,}\) we can think of \(3x\) as \(2x + x\text{,}\)
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\begin{align*} \cos{(3x)} \amp = \cos{(2x + x)}\\ \amp = \cos{(2x)} \cos{x} - \sin{(2x)} \sin{x} \end{align*}
Here, there are \(\cos{(2x)}\) and \(\sin{(2x)}\text{.}\) For \(\sin{(2x)}\text{,}\) we have \(\sin{(2x)} = 2 \sin{x} \cos{x}\text{.}\) For \(\cos{(2x)}\text{,}\) there are 3 options, and we will choose \(\cos{(2x)} = 2\cos^2{x} - 1\text{,}\) because we want the final expression to be in terms of cosine only.
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\begin{align*} \amp = \underbrace{\brac{2\cos^2{x} - 1}} \cos{x} - \underbrace{2 \sin{x} \cos{x}} \cdot \sin{x}\\ \amp = 2\cos^3{x} - \cos{x} - 2 \sin^2{x} \cos{x} \end{align*}
We will replace \(\sin^2{x} = 1 - \cos^2{x}\text{,}\) again because we want everything in terms of cosine only,
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\begin{align*} \amp = 2\cos^3{x} - \cos{x} - 2 \brac{1 - \cos^2{x}} \cos{x}\\ \amp = 2\cos^3{x} - \cos{x} - 2 \cos{x} + 2 \cos^3{x} \amp\amp \text{distributing}\\ \amp = 4\cos^3{x} - 3 \cos{x} \amp\amp \text{collecting like terms} \end{align*}
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In summary,
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\begin{gather*} \boxed{\cos{(3\theta)} = 4\cos^3{\theta} - 3 \cos{\theta}} \end{gather*}
Similarly, you can find that,
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\begin{gather*} \boxed{\sin{(3\theta)} = 3\sin{\theta} - 4 \sin^3{\theta}} \end{gather*}

Example 15.4.6. Prove the identity \(\sin{(3\theta)} = 3\sin{\theta} - 4 \sin^3{\theta}\).

Solution: We again think of \(3x\) as \(2x + x\text{,}\)
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\begin{align*} \sin(3x) \amp = \sin(2x + x)\\ \amp = \sin(2x)\cos x + \cos(2x)\sin x \end{align*}
Here, there are \(\sin(2x)\) and \(\cos(2x)\text{.}\) For \(\sin(2x)\text{,}\) we have \(\sin(2x) = 2\sin x \cos x\text{.}\) For \(\cos(2x)\text{,}\) there are 3 options, and we will choose \(\cos(2x) = 1 - 2\sin^2 x\text{,}\) because we want the final expression to be in terms of sine only.
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\begin{align*} \amp = \underbrace{2\sin x \cos x}\cos x + \underbrace{(1 - 2\sin^2 x)}\sin x\\ \amp = 2\sin x \cos^2 x + \sin x - 2\sin^3 x \end{align*}
We will replace \(\cos^2 x = 1 - \sin^2 x\text{,}\) again because we want everything in terms of sine only,
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\begin{align*} \amp = 2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\ \amp = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \amp\amp \text{distributing}\\ \amp = 3\sin x - 4\sin^3 x \amp\amp \text{collecting like terms} \end{align*}
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Similarly, we can find a tangent triple-angle identity,
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\begin{gather*} \boxed{\tan{(3\theta)} = \frac{3\tan{\theta} - \tan^3{\theta}}{1 - 3 \tan^2{\theta}}} \end{gather*}

Example 15.4.7. Prove the identity \(\tan{(3\theta)} = \frac{3\tan{\theta} - \tan^3{\theta}}{1 - 3 \tan^2{\theta}}\).

Solution: We again think of \(3x\) as \(2x + x\text{,}\)
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\begin{align*} \tan(3x) \amp = \tan(2x + x)\\ \amp = \frac{\tan(2x) + \tan x}{1 - \tan(2x)\tan x} \end{align*}
Here, there is \(\tan(2x)\text{.}\) For \(\tan(2x)\) we have the double-angle identity
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\begin{gather*} \tan(2x) = \frac{2\tan x}{1 - \tan^2 x} \end{gather*}
Then,
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\begin{align*} \tan{(3x)} \amp = \frac{\frac{2\tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2\tan x}{1 - \tan^2 x}\cdot \tan x} \end{align*}
To simplify this complex fraction, clear denominators by multiplying the numerator and denominator by \(1-\tan^2{x}\text{,}\)
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\begin{align*} \tan(3x) \amp = \frac{\frac{2\tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2\tan x}{1 - \tan^2 x}\cdot \tan x} \cdot \frac{1-\tan^2{x}}{1-\tan^2{x}}\\ \amp = \frac{2\tan x + \tan x(1 - \tan^2 x)}{(1 - \tan^2 x) - 2\tan^2 x}\\ \amp = \frac{2\tan x + \tan x - \tan^3 x}{1 - \tan^2 x - 2\tan^2 x}\\ \tan{(3x)} \amp = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \end{align*}
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