Trigonometrical Ratios and Identities 1 Question 14
14. The expression
$$ \begin{aligned} 3 \sin ^{4} \frac{3 \pi}{2}-\alpha+ & \sin ^{4}(3 \pi+\alpha) \\ & -2 \sin ^{6} \frac{\pi}{2}+\alpha+\sin ^{6}(5 \pi-\alpha) \end{aligned} $$
is equal to
(a) 0
(b) 1
(c) 3
(d) $\sin 4 \alpha+\cos 6 \alpha$
$(1986,2 M)$
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Answer:
Correct Answer: 14. (c)
Solution:
- Given expression $=$
$$ \begin{aligned} 3 \sin ^{4} & \frac{3 \pi}{2}-\alpha+\sin ^{4}(3 \pi+\alpha)-2 \sin ^{6} \frac{\pi}{2}+\alpha \\ & =3\left(\cos ^{4} \alpha+\sin ^{4} \alpha\right)-2\left(\cos ^{6} \alpha+\sin ^{6} \alpha\right) \\ & =3\left(1-2 \sin ^{2} \alpha \cos ^{2} \alpha\right)-2\left(1-3 \sin ^{2} \alpha \cos ^{2} \alpha\right) \\ & =3-6 \sin ^{2} \alpha \cos ^{2} \alpha-2+6 \sin ^{2} \alpha \cos ^{2} \alpha=1 \end{aligned} $$
