Question

The value of $3(\cos \theta -\sin \theta {)}^4+6(\sin \theta +\cos \theta {)}^2+4{\sin }^6\theta$ is

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

Answer 1

Answer: (B)

$13-4{\cos }^6\theta$

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

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

$=(1-2\sin \theta \cos \theta {)}^2+(1+2\sin \theta \cos \theta )+4{\sin }^6\theta$

$=1+4{\sin }^2\theta {\cos }^2\theta -4\sin \theta \cos \theta +1+2\sin \theta \cos \theta +4{\sin }^6\theta$

$=9+12{\cos }^{\theta }.{\sin }^2\theta +4-12{\sin }^2\theta {\cos }^2\theta +4{\cos }^6\theta$

$=13-4{\cos }^6\theta$