What is $- {sin}^{4} θ$ $-sin^4theta$ in terms of non-exponential trigonometric functions?

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What is $- {sin}^{4} θ$ $-sin^4theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-03-18T13:59:26

$= - \frac{3}{8} - \frac{cos 4 t}{8} + \frac{cos 2 t}{2}$ $= -3/8 - (cos 4t)/8 + (cos 2t)/2$

Explanation:

$2 {sin}^{2} t = (1 - cos 2 t)$ $2sin^2 t = (1 - cos 2t)$ .
Square the 2 sides:
$4 {sin}^{4} t = {(1 - cos 2 t)}^{2} = 1 + {cos}^{2} 2 t - 2 cos 2 t$ $4sin^4 t = (1 - cos 2t)^2 = 1 + cos^2 2t - 2cos 2t$
Replace ${cos}^{2} 2 t$ $cos^2 2t$ by $\frac{1 + cos 4 t}{2}$ $(1 + cos 4t)/2$
$4 {sin}^{4} t = 1 + \frac{1}{2} + \frac{cos (4 t)}{2} - 2 cos 2 t =$ $4sin^4t = 1 + 1/2 + cos (4t)/2 - 2cos 2t =$
$\frac{3}{2} + (\frac{cos (4 t)}{2}) - 2 cos 2 t$ $3/2 + (cos (4t)/2) - 2cos 2t$
Divide both sides by -4 -->

$- {sin}^{4} t = - \frac{3}{8} - \frac{cos 4 t}{8} + \frac{cos 2 t}{2}$ $- sin^4 t = - 3/8 - (cos 4t)/8 + (cos 2t)/2$

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What is $- {sin}^{4} θ$ $-sin^4theta$ in terms of non-exponential trigonometric functions?

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