Question

Sin²x ,cos2x,
Sin2x Formula

Answer 1

sin2x=2 sinx cosx
sin(square)x=1-cos(square)x
cos2x=2 cos(square)x-1
=1-2 sin(square)x

Answer 2

Answer and Explanation:

To find : Formula of $\sin^2x,\cos 2x,\sin 2x$

Solution :

The formulas of $\sin^2x$

$\sin^2 x=1-\cos^2 x$

$\sin^2 x=\frac{1-\cos 2x}{2}$

The formulas of $\sin 2x$

$\sin 2x=2\sin x\cos x$

The formulas of $\cos 2x$

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

$\cos 2x=2\cos^2 x-1$

$\cos 2x=1-2\sin^2 x$