## Trigonometry (11th Edition) Clone

$$\sin^2\frac{x}{2}=\frac{\tan x-\sin x}{2\tan x}$$
$$\sin^2\frac{x}{2}=\frac{\tan x-\sin x}{2\tan x}$$ We start with the right side $$A=\frac{\tan x-\sin x}{2\tan x}$$ $$A=\frac{\frac{\sin x}{\cos x}-\sin x}{\frac{2\sin x}{\cos x}}$$ $$A=\frac{\frac{\sin x-\sin x\cos x}{\cos x}}{\frac{2\sin x}{\cos x}}$$ $$A=\frac{(\sin x-\sin x\cos x)\cos x}{2\sin x\cos x}$$ $$A=\frac{(1-\cos x)\sin x\cos x}{2\sin x\cos x}$$ $$A=\frac{1-\cos x}{2}$$ $$A=\sin^2\frac{x}{2}$$ Since we know that $\sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$, it can be deduced that $\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$ as above. The identity is verified.