How do you express ${sin}^{4} θ - {sin}^{3} θ \cdot {cos}^{2} θ$ $sin^4theta - sin^3theta *cos^2theta$ in terms of non-exponential trigonometric functions?

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How do you express ${sin}^{4} θ - {sin}^{3} θ \cdot {cos}^{2} θ$ $sin^4theta - sin^3theta *cos^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-04-30T16:46:31

${sin}^{4} θ - {sin}^{3} θ \cdot {cos}^{2} θ$ $sin^4theta-sin^3theta*cos^2theta$

$= {sin}^{3} θ (sin θ - {cos}^{2} θ)$ $=sin^3theta(sintheta-cos^2theta)$

now
$sin 3 θ = 3 sin θ - 4 {sin}^{3} θ \Rightarrow {sin}^{3} θ = \frac{3}{4} sin θ - \frac{1}{4} sin 3 θ$ $sin3theta =3sintheta -4sin^3theta=>sin^3theta=3/4sintheta-1/4sin3theta$
and ${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$

Putting these in the main expression it becomes
$= {sin}^{3} θ (sin θ - {cos}^{2} θ)$ $=sin^3theta(sintheta-cos^2theta)$
$= (\frac{3}{4} sin θ - \frac{1}{4} sin 3 θ) (sin θ - \frac{1}{2} (1 + cos 2 θ))$ $=(3/4sintheta-1/4sin3theta)(sintheta-1/2(1+cos2theta))$
$= (\frac{3}{4} {sin}^{2} θ - \frac{1}{4} sin 3 θ sin θ - \frac{3}{8} sin θ - \frac{3}{8} sin θ cos 2 θ + \frac{1}{8} sin 3 θ - \frac{1}{4} sin 3 θ cos 2 θ)$ $=(3/4sin^2theta-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)$

$= (\frac{3}{8} (1 - cos 2 θ) - \frac{1}{4} sin 3 θ sin θ - \frac{3}{8} sin θ - \frac{3}{8} sin θ cos 2 θ + \frac{1}{8} sin 3 θ - \frac{1}{4} sin 3 θ cos 2 θ)$ $=(3/8(1-cos2theta)-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)$
$= (\frac{3}{8} (1 - cos 2 θ) - \frac{1}{8} (cos 2 θ - cos 4 θ) - \frac{3}{8} sin θ - \frac{3}{16} (sin 3 θ - sin θ) + \frac{1}{8} sin 3 θ - \frac{1}{8} (sin 5 θ - sin θ))$ $=(3/8(1-cos2theta)-1/8(cos2theta-cos4theta)-3/8sintheta-3/16(sin3theta-sintheta)+1/8sin3theta-1/8(sin5theta-sintheta))$

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How do you express ${sin}^{4} θ - {sin}^{3} θ \cdot {cos}^{2} θ$ $sin^4theta - sin^3theta *cos^2theta$ in terms of non-exponential trigonometric functions?

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