# Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$

Show that

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

Starting from the left hand side (LHS)

\begin{align} \text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\ &=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\ &=\cos^4A-\cos^2A\sin^2A+\sin^4A \end{align}

Can anyone help me to continue from here

Let $a=\cos^2 (A),b=\sin^2 (A)$ . Now use $a^3+b^3=(a+b)^3-3ab (a+b)$ also note that $a+b=1$. Hence the proof.

HINT:

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$ can also be written as

$$a^3+b^3=(a+b)^3-3ab(a+b)$$

(a+b)^3= a^3 + b^3 + 3ab(a + b)

$$=\cos^4A-\cos^2A\sin^2A+\sin^4A$$ $$=\cos^4A-2\cos^2A\sin^2A+\sin^4A + \cos^2a\sin^2a$$ $$=(\cos^2A-\sin^2A)^2 + \cos^2A\sin^2A=1-3\sin^2A\cos^2A$$ $$(\cos^2A-\sin^2A)^2 + 4\cos^2A\sin^2A=1$$ $2\cos x\sin x=\sin2x$ -> $\sin^22x=4\cos^2\sin^2x$

and

$\cos^2x-\sin^2x =\cos2x$

$$\cos^22A + sin^22A=1$$