Question

For any $\theta \in (\frac{\pi }{4},\frac{\pi }{2})$, the expression $3(\sin \theta -\cos \theta {)}^4+6(\sin \theta +\cos \theta {)}^2+4{\sin }^6\theta$ equals:

• A. $13-4{\cos }^2\theta +6{\sin }^2\theta {\cos }^2\theta$
• B. $13-4{\cos }^2\theta +6{\cos }^4\theta$
• C. $13-4{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta$
• D. $13-4{\cos }^6\theta$

Answer 1

The correct option is D $13-4{\cos }^6\theta$

$3(\sin \theta -\cos \theta {)}^4+6(\sin \theta +\cos \theta {)}^2+4{\sin }^6\theta$
$=3(1-\sin 2\theta {)}^2+6(1+\sin 2\theta )+4{\sin }^6\theta$
$=3(1+{\sin }^{2}2\theta -2\sin 2\theta )+6(1+\sin 2\theta )+4{\sin }^6\theta$
$=3+3{\sin }^{2}2\theta -6\sin 2\theta +6+6\sin 2\theta +4{\sin }^6\theta$
$=9+4{\sin }^6\theta +3{\sin }^{2}2\theta$
$=9+4-4+4{\sin }^6\theta +3\times {\left(2\sin \theta \cos \theta \right)}^2$
$=13+4({\sin }^6\theta -1)+12{\sin }^2\theta {\cos }^2\theta$
$=13+4({\sin }^2\theta -1)({\sin }^4\theta +{\sin }^2\theta +1)+12{\sin }^2\theta {\cos }^2\theta$
$=13-4{\cos }^2\theta ({\sin }^4\theta +{\sin }^2\theta +1)+12{\sin }^2\theta {\cos }^2\theta$
$=13-4{\cos }^2\theta ({\sin }^4\theta +{\sin }^2\theta +1-3{\sin }^2\theta )$
$=13-4{\cos }^2\theta ({\sin }^4\theta -2{\sin }^2\theta +1)$
$=13-4{\cos }^2\theta (1-{\sin }^2\theta {)}^2$
$=13-4{\cos }^2\theta {\cos }^4\theta$
$=13-4{\cos }^6\theta$