Question

If $$ \sin^4 \theta \cos^2 \theta = \sum_{n=0}^{\infty} a_{2n} \cos 2n\theta $$ then the least value of $ n $ for which $ a_{2n} = 0 $ is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Use trigonometric identities and expansions to simplify powers of sine and cosine into sums of cosines with multiple angles. This leads to identification of coefficients.

Answer 1

The Correct Option is D

We are expanding: \[ \sin^4 \theta \cos^2 \theta \] Use identities:
- \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \)
- \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) So: \[ \sin^4 \theta = (\sin^2 \theta)^2 = \left(\frac{1 - \cos 2\theta}{2} \right)^2 = \frac{1 - 2\cos 2\theta + \cos^2 2\theta}{4} \] Then multiply by \( \cos^2 \theta \), which is \( \frac{1 + \cos 2\theta}{2} \) So: \[ \sin^4 \theta \cos^2 \theta = \frac{1}{8}(1 - 2\cos 2\theta + \cos^2 2\theta)(1 + \cos 2\theta) \] Expand this expression — you will get terms involving \( \cos 0\theta, \cos 2\theta, \cos 4\theta, \cos 6\theta \) only. All other harmonics are eliminated due to symmetry.
Now check which coefficient becomes zero:
- It is seen that \( a_8 = 0 \), i.e. for \( 2n = 8 \Rightarrow n = 4 \)