How do you express #f(theta)=-sin^2(theta)-csc^2(theta)-sec^4theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2017-02-24T20:17:39

#f(theta)=(cos2theta+1)/2+2/(cos2theta+1)-4/(cos4theta+2cos2theta+2)#

#f(theta)=-sin^2theta-csc^2theta-sec^4theta#

#cos2theta=1-2sin^2theta#

#cos2theta=2cos^2theta-1#

#cos^2theta=1/2(cos2theta+1)#

#-sin^2theta=1/2(cos2theta+1)#

#-csc^2theta=2/(cos2theta+1)#

#sec^4theta=1/cos^4theta#

#cos^4theta=(cos^2theta)^2=(1/2(cos2theta+1))^2=#

#1/4(2cos^2 2theta+2cos2theta+1)#

#cos4theta=2cos^2 2theta-1#

#2cos^2 2theta=cos4theta+1#

#1/4(2cos^2 2theta+2cos2theta+1)=1/4(cos4theta+1+2cos2theta+1)=#
#1/4(cos4theta+2cos2theta+2)#

#-sec^4theta=-4/(cos4theta+2cos2theta+2)#

#f(theta)=(cos2theta+1)/2+2/(cos2theta+1)-4/(cos4theta+2cos2theta+2)#