Question

$\underset{n\to \infty }{lim}({\sin }^4\theta +\frac{1}{4}{\sin }^{4}2\theta +.......+\frac{1}{4^n}{\sin }^4(2^n\theta \left)\right)$ is equal to

• A. ${\sin }^2\theta$
• B. ${\sin }^4\theta$
• C. ${\cos }^2\theta$
• D. ${\cos }^4\theta$

Answer 1

The correct option is A ${\sin }^2\theta$

We know that,
${\sin }^4\theta ={\sin }^2\theta {\sin }^2\theta \Rightarrow {\sin }^4\theta ={\sin }^2\theta (1-{\cos }^2\theta )\Rightarrow {\sin }^4\theta ={\sin }^2\theta -\frac{1}{4}\left(4{\sin }^2\theta {\cos }^2\theta \right)\Rightarrow {\sin }^4\theta =({\sin }^2\theta -\frac{1}{4}{\sin }^{2}2\theta )$

Now,
$S_n=({\sin }^2\theta -\frac{1}{4}{\sin }^{2}2\theta )+(\frac{1}{4}{\sin }^{2}2\theta -\frac{1}{4^2}{\sin }^{2}4\theta )...+\left(\frac{1}{4^n}{\sin }^2\right(2^n\theta )-\frac{1}{4^{n+1}}{\sin }^2(2^{n+1}\theta \left)\right)\Rightarrow S_n={\sin }^2\theta -\frac{1}{4^{n+1}}{\sin }^2\left(2^{n+1}\theta \right)$

Therefore,
$\underset{n\to \infty }{lim}S=\underset{n\to \infty }{lim}({\sin }^2\theta -\frac{1}{4^{n+1}}{\sin }^2(2^{n+1}\theta \left)\right)={\sin }^2\theta$