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Question: \[\cos^{12}x + 3\cos^{10}x + 3\cos^{8}x + \cos^{6}x - 2\]...

cos⁡12x+3cos⁡10x+3cos⁡8x+cos⁡6x−2\cos^{12}x + 3\cos^{10}x + 3\cos^{8}x + \cos^{6}x - 2

A

0

B

1

C

3

D

sin⁡x+sin⁡2x=1\sin x + \sin^{2}x = 1

Answer

1

Explanation

Solution

=3[(−cos⁡α)4+(−sin⁡α)4]−2[cos⁡6α+sin⁡6α]3\lbrack( - \cos\alpha)^{4} + ( - \sin\alpha)^{4}\rbrack - 2\lbrack\cos^{6}\alpha + \sin^{6}\alpha\rbrack =

}{\lbrack(\cos^{2}\alpha + \sin^{2}\alpha)^{3} - 3\cos^{2}\alpha\sin^{2}\alpha}$$ $$(\cos^{2}\alpha + \sin^{2}\alpha)\rbrack$$ =$3 - 6\sin^{2}\alpha\cos^{2}\alpha - 2 + 6\sin^{2}\alpha\cos^{2}\alpha$ = 1. **Trick :** Put $\alpha = 0,\frac{\pi}{2}$; then the value of expression remains constant i.e., it is independent of α.