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April 16, 2014

April 16, 2014

Posted by **lina** on Tuesday, April 3, 2012 at 1:34pm.

(sinx)^4

- pre-calculus -
**Steve**, Tuesday, April 3, 2012 at 3:32pmsin^4

(1-cos^2)^2

(1-cos)^2(1+cos)^2

(1-cos)(1-cos)(1+cos)(1+cos)

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

so,

sin^4(x) = (1 - cos(2x))^2/4

= (1 - 2cos(2x) + cos^2(2x))/4

= (1 - 2cos(2x) + (1 - cos(4x))/2)/4

= (1 - 2cos(2x) + 1/2 - cos(4x)/2)/4

= (2 - 4cos(2x) + 1 - cos(4x))/8

= (3 - 4cos(2x) - cos(4x))/8

don't know which way you wanted to go there.

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