Double Angle Identities

Show your work for each problem.

1.
What is the double angle formula for sine?
1. a) $\sin(2\theta) = 2\sin\theta$
2. b) $\sin(2\theta) = \sin\theta + \cos\theta$
3. c) $\sin(2\theta) = 2\sin\theta\cos\theta$
4. d) $\sin(2\theta) = \sin^2\theta + \cos^2\theta$
Answer: $\sin(2\theta) = 2\sin\theta\cos\theta$
The double angle formula for sine is $\sin(2\theta) = 2\sin\theta\cos\theta$ . This comes from the sum formula $\sin(A+B)$ with $A = B = \theta$ .
2.
Which of the following is NOT a valid form of $\cos(2\theta)$ ?
1. a) $\cos(2\theta) = \cos^2\theta - \sin^2\theta$
2. b) $\cos(2\theta) = 2\cos^2\theta - 1$
3. c) $\cos(2\theta) = 1 - 2\sin^2\theta$
4. d) $\cos(2\theta) = 2\cos\theta$
Answer: $\cos(2\theta) = 2\cos\theta$
$\cos(2\theta) = 2\cos\theta$ is incorrect. The three valid forms are: $\cos^2\theta - \sin^2\theta$ , $2\cos^2\theta - 1$ , and $1 - 2\sin^2\theta$ .
3.
Calculate $\sin(60°)$ using the double angle formula with $\theta = 30°$ . Give your answer as a decimal rounded to 3 decimal places.
Answer: 0.866
$\sin(60°) = 2\sin(30°)\cos(30°) = 2 \cdot 0.5 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \approx 0.866$
4.
Evaluate $\cos^2(22.5°) - \sin^2(22.5°)$ using a double angle identity. Give your answer as a decimal rounded to 3 decimal places.
Answer: 0.707
$\cos^2(22.5°) - \sin^2(22.5°) = \cos(2 \cdot 22.5°) = \cos(45°) = \frac{\sqrt{2}}{2} \approx 0.707$
5.
If $\sin\theta = \frac{5}{13}$ and $\theta$ is in Quadrant I, what is $\sin(2\theta)$ ?
1. a) $\frac{60}{169}$
2. b) $\frac{119}{169}$
3. c) $\frac{10}{13}$
4. d) $\frac{120}{169}$
Answer: $\frac{120}{169}$
$\cos\theta = \frac{12}{13}$ (positive in Q1). Then $\sin(2\theta) = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}$ .
6.
If $\cos\theta = \frac{3}{5}$ and $\theta$ is in Quadrant IV, find $\cos(2\theta)$ . Give your answer as a fraction in the form a/b.
Answer: -7/25
$\cos(2\theta) = 2\cos^2\theta - 1 = 2 \cdot \frac{9}{25} - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25}$
7.
What is the double angle formula for tangent?
1. a) $\tan(2\theta) = \tan^2\theta - 1$
2. b) $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$
3. c) $\tan(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$
4. d) $\tan(2\theta) = 2\tan\theta$
Answer: $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$
The double angle formula for tangent is $\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$ . Note the minus sign in the denominator!
8.
Find $\sin(2\theta)$ if $\tan\theta = \frac{4}{3}$ and $\theta$ is in Quadrant I.
Answer: 24/25
1. From $\tan\theta = \frac{4}{3}$ , if opp = 4 and adj = 3, what is the hypotenuse? 5
2. What is $\sin\theta$ ? 4/5
3. What is $\cos\theta$ ? 3/5
4. Calculate $\sin(2\theta) = 2\sin\theta\cos\theta$ 24/25
9.
Derive the formula $\cos(2\theta) = 1 - 2\sin^2\theta$ from $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ .
Answer: $\cos(2\theta) = 1 - 2\sin^2\theta$
1. Start with: $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ . What identity relates $\cos^2\theta$ and $\sin^2\theta$ ? $\sin^2\theta + \cos^2\theta = 1$
2. Solve the Pythagorean identity for $\cos^2\theta$ . $\cos^2\theta = 1 - \sin^2\theta$
3. Substitute $\cos^2\theta = 1 - \sin^2\theta$ into the original formula. What do you get? $(1 - \sin^2\theta) - \sin^2\theta$
4. Simplify the expression. $1 - 2\sin^2\theta$
10.
Calculate $\tan(2\theta)$ if $\tan\theta = 2$ . Give your answer as a fraction in the form a/b.
Answer: -4/3
$\tan(2\theta) = \frac{2(2)}{1 - 2^2} = \frac{4}{1 - 4} = \frac{4}{-3} = -\frac{4}{3}$
11.
Solve $\sin(2x) = \sin(x)$ for $0 \leq x < 2\pi$ .
Answer: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$
1. Apply the double angle formula to the left side. $2\sin x\cos x = \sin x$
2. Rearrange to factor. $\sin x(2\cos x - 1) = 0$
3. What are the two cases from the factored form? sin(x) = 0 or cos(x) = 1/2
4. Find all solutions for $\sin(x) = 0$ in $[0, 2\pi)$ . 0, pi
5. Find all solutions for $\cos(x) = \frac{1}{2}$ in $[0, 2\pi)$ . pi/3, 5pi/3
12.
Find $\cos(2\theta)$ if $\sin\theta = -\frac{8}{17}$ and $\theta$ is in Quadrant III.
Answer: 161/289
1. Which form of $\cos(2\theta)$ is most convenient when you know only $\sin\theta$ ? $1 - 2\sin^2\theta$
2. What is $\sin^2\theta$ ? 64/289
3. What is $2\sin^2\theta$ ? 128/289
4. Calculate $\cos(2\theta) = 1 - 2\sin^2\theta$ . 161/289
13.
At what angle $\theta$ does the projectile range formula $R = \frac{v_0^2 \sin(2\theta)}{g}$ give maximum range?
1. a) $30°$
2. b) $45°$
3. c) $90°$
4. d) $60°$
Answer: $45°$
Maximum range occurs when $\sin(2\theta) = 1$ , which means $2\theta = 90°$ , so $\theta = 45°$ .
14.
If $\cos(2\theta) = \frac{7}{25}$ and $\theta$ is in Quadrant I, find $\cos\theta$ . Give your answer as a fraction in the form a/b.
Answer: 4/5
$2\cos^2\theta - 1 = \frac{7}{25}$ , so $2\cos^2\theta = \frac{32}{25}$ and $\cos^2\theta = \frac{16}{25}$ . In Quadrant I cosine is positive, so $\cos\theta = \frac{4}{5}$ .
15.
Use the double angle identity $\cos(2x) = 1 - 2\sin^2 x$ to express $\sin^2 x$ in terms of $\cos(2x)$ .
Answer: $\sin^2 x = \frac{1 - \cos(2x)}{2}$
1. Start from $\cos(2x) = 1 - 2\sin^2 x$ and solve for $2\sin^2 x$ . $2\sin^2 x = 1 - \cos(2x)$
2. Divide both sides by 2. What is $\sin^2 x$ ? $\sin^2 x = \frac{1 - \cos(2x)}{2}$