# How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$?

I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this: $$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$

The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$.

Can anybody help me?!

Thank you!

• So, how do you know it's an identity? Commented Sep 29, 2013 at 12:23

$$3\sin^4 x - 2\sin^6 x = 3 (\sin^2 x)^2 - 2(\sin^2x)^3 = (\sin^2 x)^2(3 - 2\sin^2 x)\cdots$$

Now we can use the Pythagorean Identity: $$\sin^2x + \cos^2x = 1 \iff \sin^2 x = 1 -\cos^2 x$$

\begin{align} 3\sin^4 x - 2\sin^6 x & = 3 (\sin^2 x)^2 - 2(\sin^2 x)^3 \\ \\ & = (\sin^2x)^2 (3 - 2\sin^2 x) \\ \\ & = (1 - \cos^2 x)^2\Big(3 - 2(1 - \cos ^2 x)\Big)\\ \\ & = (1- 2\cos^2 x + \cos^4 x)(1 + 2 \cos^2 x) \\ \\ & = 1 - 3 \cos^4 x + 2 \cos^6 x\end{align}

• needs another TU! =1 Commented Sep 29, 2013 at 23:58

L.H.S=$3\sin^4x-2\sin^6x$

=$3(\sin^2x)^2-2(\sin^2x)^3$

=$3(1-\cos^2x)^2 -2 (1-\cos^2x)^3$

=$3(1+\cos^4x -2\cos^2x) -2 (1-\cos^6x -3 \cos^2x+3\cos^4x)$

=$3+3\cos^4x -6\cos^2x -2+2\cos^6x +6\cos^2x-6\cos^4x)$

=$1-3\cos^4x+2\cos^6x$

=R.H.S

HINT:

As $\sin^2x+\cos^2x=1,$

$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x \ \ \ \ (1)$

$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$ $\implies \sin^6x+\cos^6x=1-3\sin^2x\cos^2x \ \ \ \ (2)$

Now equate the values of $\sin^2x\cos^2x$ from $(1),(2)$

Define $f(x)=3\sin^4(x)-2\sin^6(x)+3\cos^4(x)-2\cos^6(x)-1$. Show that $f'(x)=0$ (remember that $\cos^4(x)-\sin^4(x)=\cos^2(x)-\sin^2(x)$).

Michael