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July 31, 2014

July 31, 2014

Posted by **Zee** on Monday, June 24, 2013 at 11:07pm.

sin^(1/2)x*cosx - sin^(5/2)*cosx = cos^3x sq root sin x

I honestly have no clue how to approach the sin^(5/2)*cosx part of the equation

- Precal -
**Steve**, Tuesday, June 25, 2013 at 10:00amsince 5/2 = 2 + 1/2, you have

u^5/2 = u^2 * u^1/2, and so,

√sinx cosx - sin^2x √sinx cosx

√sinx cosx (1-sin^2 x)

√sinx cosx cos^2x

√sinx cos^3x

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