Question

\(\cos^2\!\left(\frac{\pi}{6}+\theta\right)-\sin^2\!\left(\frac{\pi}{6}-\theta\right)=\)

• A. \(\frac{1}{2}\cos2\theta\)
• B. \(0\)
• C. \(-\frac{1}{2}\cos2\theta\)
• D. \(\frac{1}{2}\)

Hint:

Convert squares into double-angle form for simplification.

Answer 1

The Correct Option is B

Step 1: Use identities: \[ \cos^2A-\sin^2B=\frac{1}{2}(\cos2A+\cos2B) \]
Step 2: Substitute \(A=\frac{\pi}{6}+\theta\), \(B=\frac{\pi}{6}-\theta\).
Step 3: Simplification gives zero.