Trigonometrical Ratios and Identities 1 Question 7
7. 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 (2019 Main, 9 Jan I) (a) $13-4 \cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta$
(b) $13-4 \cos ^{2} \theta+6 \cos ^{4} \theta$
(c) $13-4 \cos ^{2} \theta+6 \sin ^{2} \theta \cos ^{2} \theta$
(d) $13-4 \cos ^{6} \theta$
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Answer:
Correct Answer: 7. $\frac{1}{8}$
(c)
Solution:
- Given expression
$=3(\sin \theta-\cos \theta)^{4}+6(\sin \theta+\cos \theta)^{2}+4 \sin ^{6} \theta$ $=3\left((\sin \theta-\cos \theta)^{2}\right)^{2}+6(\sin \theta+\cos \theta)^{2}+4\left(\sin ^{2} \theta\right)^{3}$
$=3(1-\sin 2 \theta)^{2}+6(1+\sin 2 \theta)+4\left(1-\cos ^{2} \theta\right)^{3}$
$\left[\because 1+\sin 2 \theta=(\cos \theta+\sin \theta)^{2}\right.$
and $\left.1-\sin 2 \theta=(\cos \theta-\sin \theta)^{2}\right]$
$=3\left(1^{2}+\sin ^{2} 2 \theta-2 \sin 2 \theta\right)+6(1+\sin 2 \theta)$
$+4\left(1-\cos ^{6} \theta-3 \cos ^{2} \theta+3 \cos ^{4} \theta\right)$
$\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right.$
and $\left.(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}\right]$
$=3+3 \sin ^{2} 2 \theta-6 \sin 2 \theta+6+6 \sin 2 \theta+4$
$$ -4 \cos ^{6} \theta-12 \cos ^{2} \theta+12 \cos ^{4} \theta $$
$=13+3 \sin ^{2} 2 \theta-4 \cos ^{6} \theta-12 \cos ^{2} \theta+12 \cos ^{4} \theta$
$=13+3(2 \sin \theta \cos \theta)^{2}-4 \cos ^{6} \theta-12 \cos ^{2} \theta\left(1-\cos ^{2} \theta\right)$
$=13+12 \sin ^{2} \theta \cos ^{2} \theta-4 \cos ^{6} \theta-12 \cos ^{2} \theta \sin ^{2} \theta$
$=13-4 \cos ^{6} \theta$
