Question

If $p_{n} = cos^{n}\theta + sin^{n}\theta$, then $pn - p_{n-2} = kp_{n-4}$, where :

• A. $k = 1$
• B. $k = -sin^{2}\,\theta \,cos^{2}\,\theta $
• C. $k = sin^{2}\,\theta $
• D. $k = cos^{2}\,\theta $

Answer 1

Answer: (B)

$p_{n} - p_{n-2} = \left(cos^{n} \theta + sin^{n} \theta\right)-\left(cos^{n-2}\theta + sin^{n-2} \theta\right)$
$ = cos^{n-2}\theta \left(cos^{2}\theta - 1\right) + sin^{n-2} \theta \left(sin^{2} \theta - 1 \right)$
$= -sin^{2}\,\theta \,cos^{n-2}\,\theta - cos^{2}\,\theta \,sin^{n-2}\,\theta $
$= -sin^{2}\,\theta \,cos^{2}\,\theta \left(cos^{n-4}\,\theta + sin^{n-4}\,\theta\right) $
$= -sin^{2}\,\theta \,cos^{2}\,\theta p_{n-4} = kp_{n-4}$
$\Rightarrow k = -sin^{2}\,\theta \,cos^{2}\,\theta $