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What steps do we take for the following?

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

pshmath0
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user3601507
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    Hint: Do you know an identity for $1 + \tan^2\theta$? Use that, and then write the results in terms of $\sin$ and $\cos$. – paw88789 Aug 19 '14 at 10:52

3 Answers3

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

Darth Geek
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Hint

Replace $\tan(y)$ by $\frac{\sin(y)}{\cos(y)}$. Multiply numerator and denominator by $\cos^2(y)$. Simplify.

I am sure that you can take fom here.

Claude Leibovici
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Let $y=x/2$, then $$\frac{2\tan y}{1+\tan^2 y} = 2\frac{\sin y}{\cos y}\frac{1}{\sec^2 y} = 2\frac{\sin y}{\cos y}\cos^2 y=2\sin(y)\cos(y)=\sin(2y)=\sin x,$$ as required.

pshmath0
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