ISI2015-MMA-27

Question

Let $\cos ^6 \theta = a_6 \cos 6 \theta  + a_5 \cos 5 \theta + a_4 \cos 4 \theta + a_3 \cos 3 \theta + a_2 \cos 2 \theta + a_1 \cos  \theta +a_0$. Then $a_0$ is

• A. $0$
• B. $1/32$
• C. $15/32$
• D. $10/32$

Comments

Naveen Kumar 3

Question not edited properly. There is a mistake.

Answer 1

 

use this to evaluate

Answer 2

Answer: D

----------------------------------------------

$\cos^2\theta=\frac{\cos2\theta+1}{2}$ and $\cos^3\theta=\frac{\cos3\theta+3\cos\theta}{4}$.

 

Using these two identities we have:

$$\begin{align*}\cos^6\theta =&\left(\frac{\cos2\theta+1}{2}\right)^3 \\ =& \frac{\cos^3(2\theta)+1+3\cos^2(2\theta)+3\cos(2\theta)}{8}\\ =&\frac{\frac{\cos(6\theta)+3\cos(2\theta)}{4}+1+3\left(\frac{\cos(4\theta)+1}{2}\right)+3\cos(2\theta)}{8}\\ =& \text{some non-constant terms}+\frac{5}{16}\end{align*}$$

 

Hence, $\frac{5}{16}=\frac{10}{32}$.