12. Trigonometric Identities
Double Angle Identities
Multiple Choice
Given and 0 < < , find .
A
0
B
C
D
Verified step by step guidance1
Start by recalling the identity for tangent: \( \tan\theta = \frac{\sin\theta}{\cos\theta} \). Given \( \tan\theta = \frac{5}{12} \), we can set \( \sin\theta = 5k \) and \( \cos\theta = 12k \) for some positive constant \( k \).
Use the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \) to find \( k \). Substitute \( \sin\theta = 5k \) and \( \cos\theta = 12k \) into the identity: \( (5k)^2 + (12k)^2 = 1 \).
Simplify the equation: \( 25k^2 + 144k^2 = 1 \), which gives \( 169k^2 = 1 \). Solve for \( k \) to find \( k = \frac{1}{13} \).
Substitute \( k = \frac{1}{13} \) back into \( \sin\theta = 5k \) and \( \cos\theta = 12k \) to find \( \sin\theta = \frac{5}{13} \) and \( \cos\theta = \frac{12}{13} \).
Use the double angle identity for cosine: \( \cos(2\theta) = \cos^2\theta - \sin^2\theta \). Substitute \( \cos\theta = \frac{12}{13} \) and \( \sin\theta = \frac{5}{13} \) into the identity to find \( \cos(2\theta) = \left(\frac{12}{13}\right)^2 - \left(\frac{5}{13}\right)^2 \).
