How do you express #f(theta)=-sin^2(theta)-2cos^2(theta)-5sin^4theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2018-04-07T09:54:20

#f(θ)=−sin^2(θ)−2cos^2(θ)−5sin^4θ#

#cos2theta = cos^2theta - sin^2theta #
#cos2theta= 2cos^2theta - 1#
#cos2theta = 1- 2sin^2theta#

#=>(cos2theta-1)/2−(cos2theta+1)−5(sin^2θ)^2#

#=>(cos2theta-1)/2−(cos2theta+1)−5((1-cos2theta)/2)^2#

#=>(cos2theta-1)/2−(cos2theta+1)−5/4(1-cos2theta)^2#

#=>(cos2theta-1)/2−(cos2theta+1)−5/4(1+cos^2 2theta - 2cos2theta)#

#=>(cos2theta-1)/2−(cos2theta+1)−5/4(1+(cos4theta+1)/2 - 2cos2theta)#

#=>(2cos2theta-2)/4−(4cos2theta+4)/4−5/4(1+(cos4theta+1)/2 - 2cos2theta)#

#=>1/4(2cos2theta-2−4cos2theta-4−5-(5cos4theta)/2-5/2 + 10cos2theta)#

#=>1/4(8cos2theta−25/2-(5cos4theta)/2-5/2)#