Question

if sin(theta) + sin^2(theta)=1 , then prove that
cos^12(theta) + 3cos^10(theta) + 3cos^8(theta) + cos^6(theta) - 1 = 0

Answer 1

\[\sin^2(\theta)+\cos^2(\theta) =1\]\[\cos^2(\theta)=1-\sin^2(\theta)\]

\[\sin(\theta) + \sin^2(\theta)=1\]\[\sin(\theta)=1-\sin^2(\theta)\]
\[\sin(\theta)=\cos^2(\theta)\]

Answer 2

\[\cos^{12}(\theta) + 3\cos^{10}(\theta) + 3\cos^8(\theta) + \cos^6(\theta) - 1\]\[=\sin^6(\theta) + 3\sin^{5}(\theta) + 3\sin^4(\theta) + \sin^3(\theta) - 1\]