Question

$\frac{ \sqrt{1 + \sin( \frac{39\pi}{8} ) } }{ \sqrt{1 + \sin( \frac{57\pi}{8} ) } } = \tan( \frac{k\pi}{16} )$
Find the value of K. ​

Answer 1

Step-by-step explanation:

We have,

$\frac{ \sqrt{1 + \sin \bigg( \dfrac{39\pi}{8} \bigg) } }{ \sqrt{1 + \sin \bigg( \dfrac{57\pi}{8} \bigg) } } = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies\frac{ \sqrt{1 + \sin \bigg( 5\pi - \dfrac{\pi}{8} \bigg) } }{ \sqrt{1 + \sin \bigg(7\pi + \dfrac{\pi}{8} \bigg) } } = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies\frac{ \sqrt{1 + \sin \bigg( \dfrac{\pi}{8} \bigg) } }{ \sqrt{1 - \sin \bigg( \dfrac{\pi}{8} \bigg) } } = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \sqrt{\frac{ 1 + \sin \bigg( \dfrac{\pi}{8} \bigg) }{ 1 - \sin \bigg( \dfrac{\pi}{8} \bigg) } }= \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \sqrt{\frac{ 2 \cos^{2} \bigg( \dfrac{\pi}{8} \bigg) }{ 2 \sin^{2} \bigg( \dfrac{\pi}{8} \bigg) } }= \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \cot \bigg( \dfrac{\pi}{8} \bigg) = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \tan \bigg( \frac{\pi}{2} - \dfrac{\pi}{8} \bigg) = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \tan \bigg( \dfrac{3\pi}{8} \bigg) = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \tan \bigg( \dfrac{6\pi}{16} \bigg) = \tan \bigg( \frac{k\pi}{16} \bigg )$

$\implies \: k = 6$