How do you use the half-angle identity to find the exact value of csc [ (7pi) / 8 ]?

Aug 20, 2015

$\frac{2}{\sqrt{2 - \sqrt{2}}}$

Explanation:

$n = \cos \sec \setminus \frac{7 \pi}{8} = \setminus \frac{1}{\sin \setminus \frac{7 \pi}{8}} = \setminus \frac{1}{\sin \setminus \frac{\pi}{8}}$

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x = 1 - 2 {\sin}^{2} x$

$\cos \frac{\pi}{4} = 1 - 2 {\sin}^{2} \frac{\pi}{8}$

$\frac{\sqrt{2}}{2} = 1 - 2 \frac{1}{{n}^{2}}$

$\frac{2}{{n}^{2}} = 1 - \frac{\sqrt{2}}{2}$

$\frac{2}{{n}^{2}} = \frac{2 - \sqrt{2}}{2}$

${n}^{2} = \frac{4}{2 - \sqrt{2}}$

$n = \frac{2}{\sqrt{2 - \sqrt{2}}}$

Aug 20, 2015

Find $\csc \left(\frac{7 \pi}{8}\right)$

Ans: 2/(sqrt(2 - sqrt2)

Explanation:

$\csc \left(\frac{7 \pi}{8}\right) = \frac{1}{\sin} \left(\frac{7 \pi}{8}\right)$
Find $\sin \left(\frac{7 \pi}{8}\right)$. Call $\sin \left(\frac{7 \pi}{8}\right) = \sin t$
$\cos 2 t = \cos \left(\frac{14 \pi}{8}\right) = \cos \left(\frac{7 \pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
Apply the trig identity: $\cos 2 t = 1 - 2 {\sin}^{2} t$

$\cos 2 t = \frac{\sqrt{2}}{2} = 1 - 2 {\sin}^{2} t$
$2 {\sin}^{2} t = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$
${\sin}^{2} t = \frac{2 - \sqrt{2}}{4}$
$\sin \left(\frac{7 \pi}{8}\right) = \sin t = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$
Since the arc (7pi/8) is located in Quadrant II. then only the positive answer is accepted.
$\sin \left(\frac{7 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
Therefor, csc ((7pi)/8) = (2)/(sqrt(2 - sqrt2)