## Trigonometry (11th Edition) Clone

$$\sin2A\cos2A=\sin2A-4\sin^3A\cos A$$ The left side is equal to the right side, showing that the equation is an identity. The detailed steps are as below.
$$\sin2A\cos2A=\sin2A-4\sin^3A\cos A$$ We take a look at the right side first. $$X=\sin2A-4\sin^3A\cos A$$ - Now recall that $\sin2A=2\sin A\cos A$. Whether you would write $\sin2A$ here into $2\sin A\cos A$ or you realize $4\sin^3A\cos A$ can be rewritten into $(2\sin A\cos A)\times(2\sin^2A)$ is up to you... $$X=\sin2A-(2\sin A\cos A)\times(2\sin^2A)$$ $$X=\sin2A-\sin2A\times(2\sin^2A)$$ ... but eventually the point is to create 2 similar $\sin 2A$ or $2\sin A\cos A$ so that we can match them now. $$X=\sin2A(1-2\sin^2A)$$ - Finally, as in Double-Angle Identity for cosine, $1-2\sin^2A=\cos2A$. Thus, $$X=\sin2A\cos2A$$ Therefore, $$\sin2A\cos2A=\sin2A-4\sin^3A\cos A$$ The left side is equal to the right side, showing that the equation is an identity.