What is ${cos}^{4} θ + {sin}^{3} θ$ $cos^4theta+sin^3theta$ in terms of non-exponential trigonometric functions?

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What is ${cos}^{4} θ + {sin}^{3} θ$ $cos^4theta+sin^3theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-07-08T08:26:30

$= \frac{1}{8} (3 + 4 cos 2 θ + 2 cos 4 θ + 6 sin θ - 2 sin 3 θ)$ $=1/8(3+4cos2theta+2cos4 theta+6sintheta-2sin3theta)$

Explanation:

We will use the identities

$> 2 {cos}^{2} θ = (1 + cos 2 θ)$ $color(red)(>2cos^2theta=(1+cos2theta))$

$> 4 {sin}^{3} θ = 3 sin θ - sin 3 θ$ $color(blue)(>4sin^3theta=3sintheta-sin3theta)$

The given expression

$= {cos}^{4} θ + {sin}^{3} θ$ $=cos^4theta+sin^3theta$

$= \frac{1}{4} (4 {cos}^{4} θ + 4 {sin}^{3} θ)$ $=1/4(4cos^4theta+4sin^3theta)$

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

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

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

$= \frac{1}{4} (1 + 2 cos 2 θ + \frac{1}{2} (1 + cos 4 θ) + 3 sin θ - sin 3 θ)$ $=1/4(1+2cos2theta+1/2(1+cos4 theta)+3sintheta-sin3theta)$

$= \frac{1}{4} (\frac{3}{2} + 2 cos 2 θ + cos 4 θ + 3 sin θ - sin 3 θ)$ $=1/4(3/2+2cos2theta+cos4 theta+3sintheta-sin3theta)$

$= \frac{1}{8} (3 + 4 cos 2 θ + 2 cos 4 θ + 6 sin θ - 2 sin 3 θ)$ $=1/8(3+4cos2theta+2cos4 theta+6sintheta-2sin3theta)$

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What is ${cos}^{4} θ + {sin}^{3} θ$ $cos^4theta+sin^3theta$ in terms of non-exponential trigonometric functions?

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What is cos4θ+sin3θcos^4theta+sin^3theta in terms of non-exponential trigonometric functions?

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What is ${cos}^{4} θ + {sin}^{3} θ$ $cos^4theta+sin^3theta$ in terms of non-exponential trigonometric functions?