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February 1, 2015

February 1, 2015

Posted by **skye** on Thursday, August 16, 2012 at 11:24pm.

sin^4x=1/8(3-4cos2x+cos4x)

cos^3x=(1/2cosx) (1+cos2x)

thanks :)

- precalc -
**Steve**, Friday, August 17, 2012 at 12:21amcos 2x = 2cos^2 x - 1

so, 1/2 cos x (1+2cos^2 x - 1) = cos^3 x

cos 4x = 1 - 2sin^2 2x

= 1 - 8sin^2 x cos^2 x

= 1 - 8sin^2 x (1 - sin^2 x)

= 1 - 8sin^2 x + 8 sin^4 x

cos 2x = 1 - 2sin^2 x

4cos 2x = 4 - 8sin^2 x

1/8(3-4cos2x+cos4x)

= 1/8(3 - 4 + 8sin^2 x + 1 - 8sin^2 x + 8 sin^4 x)

= 1/8(8sin^4 x)

= sin^4 x

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