Answer
$$\frac{\sin2x}{2\sin x}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$
The left side is equal to the right side, so the equation is an identity.
Work Step by Step
$$\frac{\sin2x}{2\sin x}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$
1) We take a look at the left side first.
$$X=\frac{\sin2x}{2\sin x}$$
- Double-angle identity for sine: $\sin2x=2\sin x\cos x$
$$X=\frac{2\sin x\cos x}{2\sin x}$$
$$X=\cos x$$
2) Then we take a lookt at the right side.
$$Y=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$
- Half-angle identity for cosine: $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$
So, $$\cos^2\frac{x}{2}=\frac{1+\cos x}{2}$$
- Half-angle identity for sine: $\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}$
So, $$\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$$
Apply back to $Y$:
$$Y=\frac{1+\cos x}{2}-\frac{1-\cos x}{2}$$
$$Y=\frac{1+\cos x-(1-\cos x)}{2}$$
$$Y=\frac{1+\cos x-1+\cos x}{2}$$
$$Y=\frac{2\cos x}{2}=\cos x$$
3) Therefore, $$X=Y=\cos x$$
That means $$\frac{\sin2x}{2\sin x}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$
The left side is equal to the right side, so the equation is an identity.
