What is $4 {sin}^{5} θ$ $4sin^5theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2015-12-06T21:58:01

Use this identity ...

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

$4 {sin}^{5} x = 4 \times ({sin}^{2} x) \times ({sin}^{2} x) \times sin x$ $4sin^5x=4xx(sin^2x)xx(sin^2x)xxsinx$

$= 4 \times \frac{1 - cos 2 x}{2} \times \frac{1 - cos 2 x}{2} \times sin x$ $=4xx(1-cos2x)/2 xx (1-cos2x)/2 xx sinx$

$= (1 - cos 2 x) (1 - cos 2 x) (sin x)$ $=(1-cos2x)(1-cos2x)(sinx)$

Hope that helped

What is 4sin5θ4sin^5theta in terms of non-exponential trigonometric functions?