Question:
The value of $\cos ^{4} x+\sin ^{4} x-6 \cos ^{2} x \sin ^{2} x$ is
(a) cos 2x
(b) sin 2x
(c) cos 4x
(d) none of these
Solution:
(c) cos 4x
$\cos ^{4} x+\sin ^{4} x-6 \cos ^{2} x \sin ^{2} x=\cos ^{4} x+\sin ^{4} x-2 \cos ^{2} x \sin ^{2} x-4 \cos ^{2} x \sin ^{2} x$
$=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}-(2 \sin x \cos x)^{2}$
$=\cos ^{2} 2 x-\sin ^{2} 2 x$
$=\cos 4 x$