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March 30, 2015

March 30, 2015

Posted by **Natash** on Sunday, November 23, 2008 at 2:31am.

cos4x=8cos^4x-8cos^2x+1

so I'm guessing you begin with the left side cos4x=cos2(2x) then im kinda lost

- trig -
**drwls**, Sunday, November 23, 2008 at 5:34amYou began the problem correctly. Continue to use the cos "double angle" formula, applying it twice.

cos4x = cos[2(2x)] = 2 cos^2(2x) - 1

= 2(2cos^2 x - 1)^2 -1

= 2*[4 cos^4 x -4cos^2 x +1) -1

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

- trig -
**Anonymous**, Thursday, February 7, 2013 at 6:32am4cos^{2}2x-1=cos4x

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