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))#