3 [ s i n 4 ( 3 π 2 − α ) + s i n 4 ( 3 π + α ) ] - [ s i n 6 ( π 2 + α ) + s i n 6 ( 5 π − α ) ] = Options: A . &nbsp0 B . &nbsp1 C . &nbsp3 D . &nbspsin 4α + sin 6α

3[sin4(3π2−α)+sin4(3π+α)]

Question

[s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)]

[s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)]

Options:

A . 0

B . 1

C . 3

D . sin 4α + sin 6α

Answer: Option B
:
B
(b) 3

[s i n^{4} (\frac{3 π}{2} - α) + s i n^{4} (3 π + α)]

[s i n^{6} (\frac{π}{2} + α) + s i n^{6} (5 π - α)]

= 3 (- c o s α)^{4} + (- s i n α)^{4} - 2 c o s^{6} α + s i n^{6} α

= 3 (c o s^{2} α + s i n^{2} α)^{2} - 2 s i n^{2} α c o s^{2} α - 2 (c o s^{2} α + s i n^{2} α)^{3} - 3 s i n^{2} α c o s^{2} α (c o s^{2} α + s i n^{2} α)

= 3 - 6 s i n^{2} α c o s^{2} α - 2 + 6 s i n^{2} α c o s^{2} α = 3 - 2 = 1

Trick: Put

α = 0

, the value of expressions remains 1 i.e., it is independent of

α

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