Question

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

Answer: (A)

We have ,
$3( \sin \; \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4 \sin^6 \theta$
$= 3(1 - \sin^2 \theta)^2 + 6(1 + \sin^2 \theta) + 4 \sin^6 \theta$
$= 3(1 - 2 \sin^2 \theta + \sin^2 2\theta) + 6 + 6 \sin 2 \theta + 4 \sin^6 \theta$
$= 9 + 12 \; \sin^2 \theta · \cos^2 \theta + 4(1 - \cos^2 \theta)^3$
$= 13 - 4 \cos^6 \theta$