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

1 Answer
Nov 27, 2016

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))