Question

If $S_n={\cos }^n\theta +{\sin }^n\theta$ , then the value of $3S_4-2S_6$ is given by

• A. 4
• B. 0
• C. 1
• D. 7

Answer 1

Answer: (C)

Given, $S_n={\cos }^n\theta +{\sin }^n\theta$
$\therefore 3S_4-2S_6=3\left[\left({\cos }^4\theta +{\sin }^4\theta \right)\right]-2\left[{\cos }^6\theta +{\sin }^6\theta \right]$
$=3\left[{\left({\cos }^2\theta +{\sin }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta \right]$
$-2[\left({\cos }^2\theta +{\sin }^2\theta \right)({\cos }^4\theta +{\sin }^4\theta -{\cos }^2\theta {\sin }^2\theta )]$
$=3\left[1-2{\sin }^2\theta {\cos }^2\theta \right]-2[{\left({\cos }^2\theta +{\sin }^2\theta \right)}^2-3{\cos }^2\theta {\sin }^2\theta ]$
$=3-6{\sin }^2\theta {\cos }^2\theta -2+6{\cos }^2\theta {\sin }^2\theta$
$=1$