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January 31, 2015

January 31, 2015

Posted by **Terri** on Friday, June 29, 2012 at 4:25pm.

- pre cal -
**Steve**, Friday, June 29, 2012 at 4:47pmsin8x = 2sin4xcos4x

= 4sin2xcos2x(cos^2(2x)-sin^2(2x))

= 8sinx*cosx(cos^2(x)-sin^2(x))[(cos^2(x)-sin^2(x)]^2 - 4sin^2(x)cos^2(x)]

and you can expand that in many ways, one of which is, using s=sin(x) c=cos(x):

8sc^7 - 56s^3c^5 + 56s^3c^5 - 8s^7c

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