First slide

Trigonometric Identities

Question

If $\sin x + \sin^{2} x = 1$ then the value of $\cos^{12} x + 3 \cos^{10} x + 3 \cos^{8} x + \cos^{6} x - 2$ is equal to

Moderate

A

0
B

1
C

-1
D

2

Solution

We have $\sin x + \sin^{2} x = 1$
$\begin{array}{l} or \sin x = 1 - \sin^{2} x or \sin x = \cos^{2} x \\ now \cos^{12} x + 3 \cos^{10} x + 3 \cos^{8} x + \cos^{6} x - 2 \\ = \sin^{6} x + 3 \sin^{5} x + 3 \sin^{4} x + \sin^{3} x - 2 \\ = {(\sin^{2} x)}^{3} + 3 {(\sin^{2} x)}^{2} \sin x + 3 (\sin^{2} x) (\sin x)^{2} + (\sin x)^{3} - 2 \\ = {(\sin^{2} x + \sin x)}^{3} - 2 = (1)^{3} - 2 = - 1 \end{array}$

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