Question

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

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

Answer 1

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

We have,
$S=3(\sin \theta -\cos \theta {)}^4+6(\sin \theta +\cos \theta {)}^2+4{\sin }^6\theta$
$(\sin \theta -\cos \theta {)}^4=((\sin \theta -\cos \theta {)}^2{)}^2=({\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta {)}^2$
$(\sin \theta +\cos \theta {)}^2=({\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta )$
$2\sin \theta \cos \theta =\sin 2\theta$
$S=3(1-\sin 2\theta {)}^2+6(1+\sin 2\theta )+4{\sin }^6\theta$
$=3(1-2\sin 2\theta )+{\sin }^{2}2\theta )+6+6\sin 2\theta +4{\sin }^6\theta$
$=9+12{\sin }^2\theta \cdot {\cos }^2\theta +4(1-{\cos }^2\theta {)}^3$
$(1={\cos }^2\theta {)}^3=1-{\cos }^6\theta -3{\cos }^2\theta (1-{\cos }^2\theta )$
$=1-{\cos }^6\theta -3{\cos }^2\theta {\sin }^2\theta$
$S=9+12{\sin }^2\theta \cdot {\cos }^2\theta +4(1-{\cos }^6\theta -3{\cos }^2\theta {\sin }^2\theta )$
$=13-4{\cos }^6\theta$.