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Question #78bc0

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Mar 1, 2017

$6 {sin}^{4} x$ $6sin^4x$

$= \frac{6}{4} \times {(2 {sin}^{2} x)}^{2}$ $=6/4xx(2sin^2x)^2$

$= \frac{3}{2} \times {(1 - cos 2 x)}^{2}$ $=3/2xx(1-cos2x)^2$

$= \frac{3}{2} \times (1 - 2 cos 2 x + {cos}^{2} 2 x)$ $=3/2xx(1-2cos2x+cos^2 2x)$

$= \frac{3}{2} \times (1 - 2 cos 2 x + \frac{1}{2} (1 + cos 4 x))$ $=3/2xx(1-2cos2x+1/2(1+cos4x))$

$= \frac{3}{4} \times (3 - 4 cos 2 x + cos 4 x))$ $=3/4xx(3-4cos2x+cos4x))$

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