<s>[INST] How To Tackle Trigonometric Proofs involving $4$th and $6$th powers? 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.
 [/INST] 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 $$LHS=\cos^6 A+\sin^6 A$$ $$=(\cos^2 A)^3+(\sin^2 A)^3$$ $$=(\cos^2A+\sin^2A)(\cos^4 A+\sin^4 A-\cos^2A\sin^2A)$$ $$=(1)(\cos^4 A+\sin^4 A+2\cos^2A\sin^2A-3\cos^2A\sin^2A)$$ $$=(\cos^2A+\sin^2 A)^2-3\sin^2A\cos^2A$$ $$=(1)^2-\sin^2A3\cos^2A$$ $$=\color{blue}{1-3\sin^2 A\cos^2A=RHS}$$
</s>