How To Tackle Trigonometric Proofs involving $4$th and $6$th powers?

Question

How do I prove that

$\cos^4A - \sin^4A+1=2\cos^2A$

$\cos^6A + \sin^6A =1-3\sin^2A\cdot\cos^2A$

I was going through a very old and very rich book of Plane Trigonometry to build a nice foundation for calculus , when I had came across these two rather interesting equations.

Unfortunately the book i'm reading never mentions how to imperatively solve such a proof, and oddly enough I was never exposed on how to prove these types of equations when I took precalc last semester as well.

Hope you guys can help me out.

Answer 1

Note that $\color{Green}{\cos^2x+\sin^2x=1}$ and $\color{Green}{\cos^2x-\sin^2x=\cos2x}.$ $$\cos^4x-\sin^4x=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)$$ $$\cos^6x+\sin^6x=(\cos^2x+\sin^2x)^3-3\sin^2x\cos^2x(\cos^2x+\sin^2x)$$

Answer 2

1: Let's take $$LHS=\cos^4 A-\sin^4 A+1$$ $$=\cos^4 A+(1-\sin^4 A)$$ $$=\cos^4 A+(1-\sin^2 A) (1+\sin^2 A)$$ $$=\cos^4 A+\cos^2 A (1+\sin^2 A)$$ $$=\cos^2 A(\cos^2A+\sin^2 A+1)$$ $$=\cos^2 A(1+1)$$ $$=\color{blue}{2\cos^2 A=RHS}$$ 2: Again let's take $$LHS=\cos^6 A+\sin^6 A$$ $$=(\cos^2 A)^3+(\sin^2 A)^3$$ $$=(\cos^2A+\sin^2A)(\cos^4 A+\sin^4 A-\cos^2A\sin^2A)$$ $$=(1)(\cos^4 A+\sin^4 A+2\cos^2A\sin^2A-3\cos^2A\sin^2A)$$ $$=(\cos^2A+\sin^2 A)^2-3\sin^2A\cos^2A$$ $$=(1)^2-\sin^2A3\cos^2A$$ $$=\color{blue}{1-3\sin^2 A\cos^2A=RHS}$$