How do you find sin (x/2), given cos x = -5/8, with pi/2 < x <pi?

May 18, 2015

Recall the formula for a cosine of a sum of two angles:
$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cdot \cos \left(\beta\right) - \sin \left(\alpha\right) \cdot \sin \left(\beta\right)$

Use it for $\alpha = \beta = \frac{x}{2}$:
$\cos \left(\frac{x}{2} + \frac{x}{2}\right) = {\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)$

Substitute $\frac{x}{2} + \frac{x}{2} = x$
Use the identity ${\sin}^{2} \left(\gamma\right) + {\cos}^{2} \left(\gamma\right) = 1$ for $\gamma = \frac{x}{2}$
The result is:
$\cos \left(x\right) = 1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)$

Now we can use the value of $\cos \left(x\right) = - \frac{5}{8}$.
$1 - 2 {\sin}^{2} \left(\frac{x}{2}\right) = - \frac{5}{8}$
${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1 + \frac{5}{8}}{2} = \frac{13}{16}$
$\sin \left(\frac{x}{2}\right) = \pm \frac{\sqrt{13}}{4}$

If $\frac{\pi}{2} < x < \pi$, $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$. For these angles function sine is positive.
Therefore, $\sin \left(\frac{x}{2}\right) = \frac{\sqrt{13}}{4}$.