# 2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to

Question:

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Solution:

The given expression is $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$.

Simplifying the given expression, we have

$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$

$=2 \sin ^{6} \theta+2 \cos ^{6} \theta-3 \sin ^{4} \theta-3 \cos ^{4} \theta$

$=\left(2 \sin ^{6} \theta-3 \sin ^{4} \theta\right)+\left(2 \cos ^{6} \theta-3 \cos ^{4} \theta\right)$

$=\sin ^{4} \theta\left(2 \sin ^{2} \theta-3\right)+\cos ^{4} \theta\left(2 \cos ^{2} \theta-3\right)$

$=\sin ^{4} \theta\left\{2\left(1-\cos ^{2} \theta\right)-3\right\}+\cos ^{4} \theta\left\{2\left(1-\sin ^{2} \theta\right)-3\right\}$

$=\sin ^{4} \theta\left(2-2 \cos ^{2} \theta-3\right)+\cos ^{4} \theta\left(2-2 \sin ^{2} \theta-3\right)$

$=\sin ^{4} \theta\left(-1-2 \cos ^{2} \theta\right)+\cos ^{4} \theta\left(-1-2 \sin ^{2} \theta\right)$

$=-\sin ^{4} \theta-2 \sin ^{4} \theta \cos ^{2} \theta-\cos ^{4} \theta-2 \cos ^{4} \theta \sin ^{2} \theta$

$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{4} \theta \sin ^{2} \theta-2 \sin ^{4} \theta \cos ^{2} \theta$

$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta(1)$

$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta$

$=-\left(\sin ^{4} \theta+\cos ^{4} \theta+2 \cos ^{2} \theta \sin ^{2} \theta\right)$

$=-\left\{\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta\right\}$

$=-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}$

$=-(1)^{2}$

$=-1$

Therefore, the correct option is (c).