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

The correct option is C

$1$

Explanation for the correct option.

Given, $S_n={\cos }^n\theta +{\sin }^n\theta$.

$\begin{array}{rcl}3S_4-2S_6& =& 3\left[{\cos }^4\theta +{\sin }^4\theta \right]-2\left[{\cos }^6\theta +{\sin }^6\theta \right]\\ & =& 3\left[{\left({\cos }^2\theta \right)}^2+{\left({\sin }^2\theta \right)}^2\right]-2\left[{\left({\cos }^2\theta \right)}^3+{\left({\sin }^2\theta \right)}^3\right]\\ & =& 3\left[\left({\cos }^2\theta +{\sin }^2\theta \right)-2{\sin }^2\theta {\cos }^2\theta \right]-2\left[{\left({\cos }^2\theta +{\sin }^2\theta \right)}^3-3{\sin }^2\theta {\cos }^2\theta \left({\cos }^2\theta +{\sin }^2\theta \right)\right]\\ & =& 3\left[1-\frac{1}{2}{\left(2\sin \theta \cos \theta \right)}^2\right]-2\left[1-\frac{3}{4}{\left(2\sin \theta \cos \theta \right)}^2\right]\\ & =& 3\left[1-\frac{1}{2}{\left(\sin 2\theta \right)}^2\right]-2\left[1-\frac{3}{4}{\left(\sin 2\theta \right)}^2\right]\left[2\sin \theta \cos \theta =\sin 2\theta \right]\\ & =& 3-2\\ & =& 1\end{array}$

Hence, option C is correct.