Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $

Question

$$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$ I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way. I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$ And I am stuck...

Answer 1

In your last formula

• Your nominator is $\sin x$ due to $\sin 2x = 2\sin x \cos x$, written for $\frac{x}{2}$.
• Your denominator is $1$, due to $\sin^2 x + \cos^2 x = 1$.

Answer 2

$\sin(2x) = 2\sin(x)\cos(x)$

$\sin(2x) =\dfrac{2\sin(x)\cos(x)}{\sin^2(x)+\cos^2(x)}$

divide throughout by $\cos^2(x)$

$\sin(2x) = \dfrac{2\tan(x)}{1+\tan^2(x)}$

let $x\to\frac x2$ to get the desired answer

Answer 3

From here

$$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$

divide by $\cos^2 \frac{\alpha}{2} $.

Answer 4

$$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} =\frac {\frac{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos^2\frac{\alpha}2}}{\frac{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}}{\cos^2\frac{\alpha}2}}=\frac{2\tan\frac{\alpha}2}{\tan^2\frac{\alpha}2+1}$$

Answer 5

$$\sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\frac{2\tan\dfrac{\alpha}{2}}{\sec^2 \dfrac{\alpha}{2}}=\frac{2\tan\dfrac{\alpha}{2}}{1+\tan^2 \dfrac{\alpha}{2}}$$