How do you express ${cos}^{3} θ + {sec}^{2} θ$ $cos^3theta+sec^2theta$ in terms of non-exponential trigonometric functions?

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How do you express ${cos}^{3} θ + {sec}^{2} θ$ $cos^3theta+sec^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-11-27T17:23:27

${cos}^{3} θ + {sec}^{2} θ$ $cos^3theta+sec^2theta$

$= \frac{1}{4} (cos 3 θ + 3 cos θ) + \frac{2}{2 {cos}^{2} θ}$ $=1/4(cos3theta+3costheta)+2/(2cos^2theta)$

$= \frac{1}{4} (cos 3 θ + 3 cos θ) + \frac{2}{1 + cos 2 θ}$ $=1/4(cos3theta+3costheta)+2/(1+cos2theta)$

$= \frac{cos 3 θ + cos θ + cos 3 θ cos 2 θ + 3 cos θ cos 2 θ + 8}{4 (1 + cos 2 θ)}$ $=(cos3theta+costheta+cos3thetacos2theta+3costhetacos2theta+8)/(4(1+cos2theta))$

$= \frac{2 cos 3 θ + 2 cos θ + 2 cos 3 θ cos 2 θ + 6 cos θ cos 2 θ + 16}{8 (1 + cos 2 θ)}$ $=(2cos3theta+2costheta+2cos3thetacos2theta+6costhetacos2theta+16)/(8(1+cos2theta))$

$= \frac{2 cos 3 θ + 2 cos θ + cos 5 θ + cos θ + 3 cos θ + 3 cos 3 θ + 16}{8 (1 + cos 2 θ)}$ $=(2cos3theta+2costheta+cos5theta+costheta+3costheta+3cos3theta+16)/(8(1+cos2theta))$

$= \frac{5 cos 3 θ + 6 cos θ + cos 5 θ + 16}{8 (1 + cos 2 θ)}$ $=(5cos3theta+6costheta+cos5theta+16)/(8(1+cos2theta))$

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How do you express ${cos}^{3} θ + {sec}^{2} θ$ $cos^3theta+sec^2theta$ in terms of non-exponential trigonometric functions?

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How do you express cos^3theta+sec^2theta cos3θ+sec2θcos^3theta+sec^2theta in terms of non-exponential trigonometric functions?

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How do you express ${cos}^{3} θ + {sec}^{2} θ$ $cos^3theta+sec^2theta$ in terms of non-exponential trigonometric functions?