Trigonometrical Ratios And Identities Ques 16
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) $\sin 4 \alpha+\cos 6 \alpha$
$(1986,2 M)$
Show Answer
Answer:
Correct Answer: 16.(b)
Solution:
Formula:
Periodicity Identities (in Radians):
- 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} $
BETA
