Question

Find the exact values of sin0/2 and cos0/2 for sin 0=11/19 on the interval 0 < 0 < 90

Answer 1

Step-by-step explanation:

$\sin( \frac{ \alpha }{2} ) = \sqrt{ \frac{1 - \cos( \alpha ) }{2} }$

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{1 + \cos( \alpha ) }{2} }$

Find cos using Pythagorean theorem.

$( \sin( \alpha ) ) {}^{2} + ( \cos( \alpha ) ) {}^{2} = 1$

$( \frac{11}{19} ) { }^{2} + ( \cos( \alpha ) ) {}^{2} = 1$

$( \frac{121}{281} ) + ( \cos( \alpha ) ) {}^{2} = 1$

$( \cos( \alpha ) ) {}^{2} = \frac{160}{281}$

$\cos( \alpha ) = \frac{4 \sqrt{10} }{19}$

Now, we use the formula

$\sin( \frac{ \alpha }{2} ) = \sqrt{ \frac{1 - \frac{4 \sqrt{10} }{19} }{2} }$

$\sin( \frac{ \alpha }{2} ) = \sqrt{ \frac{ \frac{19 - 4 \sqrt{10} }{19} }{2} }$

$\sin( \frac{ \alpha }{2} ) = \sqrt{ \frac{19 - 4 \sqrt{10} }{38} }$

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{1 + \cos( \alpha ) }{2} }$

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{19 + 4 \sqrt{10} }{38} }$