# Forum JEE Advanced (IIT) 12th Pass(Dropper) Sin theta + sin^2theta + sin ^…

Rohit Mittal

`Sin theta + sin^2theta + sin ^3theta = 1 prove that cos^6theta - 4cos^4theta + 8cos^2theta=4`

11 months ago

## Answers (1)

pooja singh

sinA + sin^2A + sin^3A = 1

=> sinA + sin^3A = 1 - sin^2A

=> sinA ( 1+sin^2A ) = cos?^2A

=> sinA ( 1 + 1 - cos^2A ) = cos^2A

=> sinA ( 2 - cos^2A ) = cos^2A

=> sinA = cos^2A / ( 2 - cos^2A)

=> sqrt ( 1 - cos^2A ) = cos^2A / 2- cos^2A

=> 1 - cos^2A = ( cos^2A )2 / ( 2 - cos^2A )2

=> 1 - cos^2A = cos^4A / ( 4 + cos^4A - 4 cos^2A )

=> ( 1 - cos^2A ) ( 4 + cos^4A - 4 cos^2A ) = cos^4A

=> ( ?4 + cos^4A - 4 cos^2A -4cos?^2A - cos^6A + 4cos^4A = cos^4A

=> ( 4 -8cos^2A - cos^6A + 4cos^4A ) = cos^4A - cos^4A

=> ( 4 -8cos^2A - cos^6A + 4 cos^4A ) = 0

=>** 4 = 8cos^2A + cos^6A - 4cos^4A**

11 months ago

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