Question

How do I factor the algebraic expression below in terms of a single trigonometric function?

cos x - sin 2x - 1

Answer 1

`\cos x-\sin^2x-1\\/use\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x/\\\\=\cos x-(1-\cos^2x)-1=\cos x-1+\cos^2x-1\\=\cos x-1+\cos^2x-1^2\\/use\ a^2-b^2=(a-b)(a+b)/\\\\=\cos x-1+(\cos x-1)(\cos x+1)\\=1(\cos x-1)+(\cos x+1)(\cos x-1)\\\\=\boxed{(\cos x-1)(1+\cos x+1)=(\cos x-1)(\cos x+2)}`