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

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Answer 1 · 2016-11-27T17:23:27

$cos^3theta+sec^2theta$

$=1/4(cos3theta+3costheta)+2/(2cos^2theta)$

$=1/4(cos3theta+3costheta)+2/(1+cos2theta)$

$=(cos3theta+costheta+cos3thetacos2theta+3costhetacos2theta+8)/(4(1+cos2theta))$

$=(2cos3theta+2costheta+2cos3thetacos2theta+6costhetacos2theta+16)/(8(1+cos2theta))$

$=(2cos3theta+2costheta+cos5theta+costheta+3costheta+3cos3theta+16)/(8(1+cos2theta))$

$=(5cos3theta+6costheta+cos5theta+16)/(8(1+cos2theta))$

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