# If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$

So as the title states I've got the following problem:

If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$

So I'd guess that this probably involves the formula for half-angles, but that is is a dead-end.

Any suggestions?

$$\frac{2\tan\left(\frac{x}{2} \right)}{1+\tan^2\left(\frac{x}{2} \right)}=$$ We know that $\tan^2(x)+1=\sec^2(x)$, so: $$\frac{2\tan\left(\frac{x}{2} \right)}{\sec^2\left(\frac{x}{2} \right)}=$$ $$\frac{2\tan\left(\frac{x}{2} \right)}{\frac{1}{\cos^2\left(\frac{x}{2} \right)}}=$$ $$2\tan\left(\frac{x}{2} \right)\cos^2\left(\frac{x}{2} \right)=$$ $$2\frac{\sin\left(\frac{x}{2} \right)}{\cos\left(\frac{x}{2} \right)}\cos^2\left(\frac{x}{2} \right)=$$ $$2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)=$$ We know that $\sin(2x)=2\sin(x)\cos(x)$, so: $$\sin\left(2\frac{x}{2} \right)=\sin(x)$$

• Thanks @abiessu! Dec 7, 2017 at 20:02
• 'Cheers. First time however i'm seeing the the halv-angle of tan expressed in terms of tan :) Dec 7, 2017 at 20:08

$$\sin x = \overbrace{\sin\left( 2\cdot \frac x 2 \right) = 2\sin\left( \frac x 2 \right) \cos\left( \frac x 2 \right)}^{\large\text{the double-angle formula for the sine}}$$

If $z = \tan \frac x 2,$ then what are $\sin \frac x 2$ and $\cos \frac x 2\,$?

In the first quadrant, we can say $\tan = \dfrac \sin \cos$ and $\sin^2+\cos^2 = 1,$ and thereby conclude that $$\sin = \frac \tan {\sqrt{1+\tan^2}} \text{ and } \cos = \frac 1 {\sqrt{1+\tan^2}}$$ so you have $$2\cdot \frac z {1+z^2} \cdot \frac 1 {\sqrt{1+z^2}} = \frac {2z}{1+z^2}.$$ In the fourth quadrant, the denominator will be the same, but the sign of the tangent will agree with the sign of the sine.

And in the other two quadrants, the whole function just repeats.

• Thank you. But i'm really sure how you get the relationship: $\sin = \frac \tan {\sqrt{1+\tan^2}} \text{ and } \cos = \frac 1 {\sqrt{1+\tan^2}}$ Dec 7, 2017 at 20:31