Question

The value of $3{\left(\cos \theta -\sin \theta \right)}^4+6{\left(\sin \theta +\cos \theta \right)}^2+4{\sin }^6\theta$ is, where $\theta \in (\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2})$

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

Answer 1

The correct option is B

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

Explanation for the correct answer:

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

$\Rightarrow$ $S=3{\left({\left(\cos \theta -\sin \theta \right)}^2\right)}^2+6{\left(\sin \theta +\cos \theta \right)}^2+4{\left({\sin }^2\theta \right)}^3$ $\left[\because {\left(a-b\right)}^2=a^2+b^2-2ab\right]$

$\Rightarrow$ $S=3{\left(\left({\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta \right)\right)}^2+6\left({\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta \right)+4{\left(1-{\cos }^2\theta \right)}^3$

$\Rightarrow$ $S=3{\left(1-\sin 2\theta \right)}^2+6\left(1+\sin 2\theta \right)+4\left(1-{\cos }^6\theta -3{\cos }^2\theta (1-{\cos }^2\theta \right))$ $\left[\because \sin 2\theta =2\sin \theta \cos \theta ,{\sin }^2\theta +{\cos }^2\theta =1\right]$

$\Rightarrow$ $S=3\left(1+{\sin }^{2}2\theta -2\sin 2\theta \right)+6\left(1+\sin 2\theta \right)+4\left(1-{\cos }^6\theta -3{\cos }^2\theta {\sin }^2\theta \right))$

$\Rightarrow$ $S=3+3{\sin }^{2}2\theta -6\sin 2\theta +6+6\sin 2\theta +4-4{\cos }^6\theta -12{\sin }^2\theta {\cos }^2\theta$

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

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

Therefore, the value of the expression $3{\left(\cos \theta -\sin \theta \right)}^4+6{\left(\sin \theta +\cos \theta \right)}^2+4{\sin }^6\theta$ is $13-4{\cos }^6\theta$.

Hence, option $\left(B\right)$ is the correct answer.