Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$

Question

Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$

There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$

By trigonometry, we know the ratio of the triangle, that is the opposite to adjacent is $2t : 1-t^2$, and hence it follows the hypotenuse is $\sqrt{(2t)^2+(1-t^2)^2} = \sqrt{(t^2+1)^2} = (t^2+1)$

My question is, can the above square rooted answer be $-(t^2+1)$ also? Why do we reject the negative answer?

Answer 1

Because $$\frac{2t}{1+t^2}=\frac{2\tan\frac{\theta}{2}}{\frac{1}{\cos^2\frac{\theta}{2}}}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\sin\theta.$$ You can use your formula, but we need to write before: $$\frac{1-t^2}{1+t^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=\cos\theta.$$ Id est, $$\sin\theta=\tan\theta\cos\theta=\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}=\frac{2t}{1+t^2}.$$

Answer 2

Usually, when you write $\sqrt n$, you mean the positive root. This is why we write $\pm$ in the quadratic formula $-b\pm\sqrt{b^2-4ac}\over2a$. It shows that we have to take into consideration both square roots, the positive and the negative.

As another method of understanding, if you get the value as negative, then you have a negative distance between two points, which is impossible.

Answer 3

I think the second "solution" comes from squaring then unsquaring. This process can often add solutions to an equation. Using a much simpler example:

$$ \begin{align} & &x-1 &= 2 \\ &\Rightarrow &(x-1)^2 &= 4 \\ &\Rightarrow &x^2-2x+1 &= 4 \\ &\Rightarrow &x^2-2x-3 &= 0 \\ &\Rightarrow &(x-3)(x+1) &= 0 \\ &\Rightarrow & x &=3,-1 \end{align} $$

Whenever you square an equation in order to solve it, you should check all the "solutions" you come out with.

In my example, using the first line to check both "solutions", obviously $x=-1$ is rubbish.

In your example, checking both "solutions" $\sin\theta = \frac{2t}{1+t^2}$ and $\sin\theta = \frac{2t}{-(1+t^2)}$ is slightly more difficult. By considering the graphs of $y=\tan\frac{\theta}{2}$ and $y=\sin\theta$ we can see that $$\sin\theta>0 \iff t=\tan\frac{\theta}{2}>0 $$ This is true for $\sin\theta = \frac{2t}{1+t^2}$ but not for $\sin\theta = \frac{2t}{-(1+t^2)}$, so we discard the second solution.

Answer 4

No, because $$ \sqrt {x^2} = |x|$$

Thus $$ \sqrt{(t^2+1)^2} = |(t^2+1)|=(t^2+1)$$