Asked by Mintz
How would I express sin^6 theta in multiples of cos theta?
Thanks
Thanks
Answers
Answered by
MathMate
sin^6 theta
=(1-cos²(θ))³
=1-3cos²(θ)+3cos^4(θ)-cos^6(θ)
=(1-cos²(θ))³
=1-3cos²(θ)+3cos^4(θ)-cos^6(θ)
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