## Algebra 2 (1st Edition)

$2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$
The half -angle formula can be defined as: $\tan (\dfrac{x}{2})=\dfrac{1- \cos x}{\sin x}$ Now, we have $2 \sin^2 x \tan (\dfrac{x}{2})=2 \sin^2 x \times \dfrac{1- \cos x}{\sin x}$ or, $=\dfrac{2 \sin^2 x-2 \sin^2 x \cos x}{\sin x}$ or, $=2 \sin x - 2\sin x \cos x$ or, $= 2\sin x - \sin 2x$ Thus, it has been proven that $2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$