How do you express cos^3theta+sec^2theta cos3θ+sec2θ in terms of non-exponential trigonometric functions?

1 Answer
Nov 27, 2016

cos^3theta+sec^2thetacos3θ+sec2θ

=1/4(cos3theta+3costheta)+2/(2cos^2theta)=14(cos3θ+3cosθ)+22cos2θ

=1/4(cos3theta+3costheta)+2/(1+cos2theta)=14(cos3θ+3cosθ)+21+cos2θ

=(cos3theta+costheta+cos3thetacos2theta+3costhetacos2theta+8)/(4(1+cos2theta))=cos3θ+cosθ+cos3θcos2θ+3cosθcos2θ+84(1+cos2θ)

=(2cos3theta+2costheta+2cos3thetacos2theta+6costhetacos2theta+16)/(8(1+cos2theta))=2cos3θ+2cosθ+2cos3θcos2θ+6cosθcos2θ+168(1+cos2θ)

=(2cos3theta+2costheta+cos5theta+costheta+3costheta+3cos3theta+16)/(8(1+cos2theta))=2cos3θ+2cosθ+cos5θ+cosθ+3cosθ+3cos3θ+168(1+cos2θ)

=(5cos3theta+6costheta+cos5theta+16)/(8(1+cos2theta))=5cos3θ+6cosθ+cos5θ+168(1+cos2θ)