Trigonometry (11th Edition) Clone

$$\frac{\sin2x}{2\sin x}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$ The left side is equal to the right side, so the equation is an identity.
$$\frac{\sin2x}{2\sin x}=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$ 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.