Question

If \[ \sin^4\theta \cos^2\theta=\sum_{n=0}^{\infty} a_{2n}\cos 2n\theta, \] then the least \(n\) for which \[ a_{2n}=0 \] is

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

While expanding trigonometric powers, use: \[ \sin^2\theta=\frac{1-\cos 2\theta}{2} \] and \[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \] to convert everything into cosine multiple-angle terms.

Answer 1

The Correct Option is D

Step 1: Express powers in terms of cosine multiples.
We have \[ \sin^4\theta \cos^2\theta \] Using \[ \sin^2\theta=\frac{1-\cos 2\theta}{2} \] and \[ \cos^2\theta=\frac{1+\cos 2\theta}{2} \] So, \[ \sin^4\theta=\left(\frac{1-\cos 2\theta}{2}\right)^2 \] Hence, \[ \sin^4\theta \cos^2\theta = \left(\frac{1-\cos 2\theta}{2}\right)^2 \left(\frac{1+\cos 2\theta}{2}\right) \]

Step 2: Simplify the expression.
\[ = \frac{(1-2\cos 2\theta+\cos^2 2\theta)(1+\cos 2\theta)}{8} \] Multiplying, \[ = \frac{1-\cos 2\theta-\cos^2 2\theta+\cos^3 2\theta}{8} \]

Step 3: Convert higher powers into multiple angles.
Using \[ \cos^2 2\theta=\frac{1+\cos 4\theta}{2} \] and \[ \cos^3 x=\frac{3\cos x+\cos 3x}{4} \] Therefore, \[ \cos^3 2\theta= \frac{3\cos 2\theta+\cos 6\theta}{4} \] Substituting, \[ \sin^4\theta \cos^2\theta = \frac{1-\cos 2\theta-\frac{1+\cos 4\theta}{2} +\frac{3\cos 2\theta+\cos 6\theta}{4}}{8} \]

Step 4: Simplify coefficients.
Taking LCM inside the bracket, \[ = \frac{ 4-4\cos 2\theta-2-2\cos 4\theta +3\cos 2\theta+\cos 6\theta }{32} \] \[ = \frac{ 2-\cos 2\theta-2\cos 4\theta+\cos 6\theta }{32} \] Thus, \[ \sin^4\theta \cos^2\theta = \frac{1}{16} -\frac{1}{32}\cos 2\theta -\frac{1}{16}\cos 4\theta +\frac{1}{32}\cos 6\theta \]

Step 5: Identify the coefficients \(a_{2n}\).
Comparing with \[ \sum_{n=0}^{\infty} a_{2n}\cos 2n\theta, \] we get: \[ a_0=\frac{1}{16}, \quad a_2=-\frac{1}{32}, \quad a_4=-\frac{1}{16}, \quad a_6=\frac{1}{32} \] There is no \(\cos 8\theta\) term. Hence, \[ a_8=0 \] Since \[ 2n=8, \] we get \[ n=4 \]

Step 6: Final conclusion.
Therefore, the least value of \(n\) for which \[ a_{2n}=0 \] is \[ \boxed{4} \]