Given cos(x) = -3/-7 and pi/2 < x < pi, how do you find cos(x/2)?

Jul 27, 2018

$x = 1.13$

Explanation:

$\frac{3}{7} = \cos x$

$= \cos \left(2 \times \frac{x}{2}\right)$

$= {\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)$

$= {\cos}^{2} \left(\frac{x}{2}\right) - \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right)$

$= 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1$

$\frac{3}{7} = 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1$

$2 {\cos}^{2} \left(\frac{x}{2}\right) = \frac{10}{7}$

${\cos}^{2} \left(\frac{x}{2}\right) = \frac{5}{7}$

$\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{5}{7}}$

The domain was originally $\frac{\pi}{2} < x < \pi$ but since it is now $\frac{x}{2}$, the domain has changed to $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ ie first quadrant only

Therefore, $\cos \left(\frac{x}{2}\right) = \sqrt{\frac{5}{7}}$ only as cosine is positive in the first and fourth quadrant and negative in the second and third quadrant

$\frac{x}{2} = 0.56$

$x = 1.13$

Jul 27, 2018

$\cos x = - \frac{3}{- 7} = \frac{3}{7} \Rightarrow x \in {Q}_{1}$ or ${Q}_{4}$.

So, $\frac{\pi}{2} < x < \pi$ is incompatible.

If $\cos x = - \frac{3}{7} , x \in {Q}_{2}$ or ${Q}_{3} \mathmr{and} \frac{x}{2} \in {Q}_{1}$or Q_2#.

$\cos \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 + \cos x\right)} = \sqrt{\frac{2}{7}} , \mathmr{if} x \in {Q}_{2}$, and
$= - \sqrt{\frac{2}{7}} , x \in {Q}_{3}$.

If $x = \frac{2}{3} \pi , \cos x = - \frac{1}{2}$ and $\cos \left(\frac{x}{2}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
If $x = \frac{4}{3} \pi , \cos x = - \frac{1}{2}$ and
$\cos \left(\frac{x}{2}\right) = \cos \left(\frac{2}{3} \pi\right) = - \frac{1}{2}$