Find $\tan \left(\frac{\theta}{2}\right)$, given $\sin (\theta) = \frac 35$, with $90^\circ < \theta < 180^\circ$

Question

Find tan theta/2, given sin theta = 3/5, with 90^∘ < theta < 180^∘. I don't know how to solve it! Help? do I use the tangent half-identity formula?

Answer 1

Recall that$$\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}$$

$$90^{\circ}<\theta<180^{\circ}\implies\theta\in\text{Quadrant}\space II$$

If we know that $\sin\theta=\frac{3}{5}$ in Quadrant $II$, then drawing the appropriate triangle in Quadrant $II$ will show that $$\cos\theta=-\frac{4}{5}$$

It helps to notice that this is a $3-4-5$ triangle. Now that we know $\sin\theta$ and $\cos\theta$ as required by the original identity for $\tan\left(\frac{\theta}{2}\right)$, we can plug those values into the identity.