How do you express f(theta)=-sin^2(theta)-2cos^2(theta)-5sin^4thetaf(θ)=sin2(θ)2cos2(θ)5sin4θ in terms of non-exponential trigonometric functions?

1 Answer
Apr 7, 2018

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)