Question

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

Answer 1

Explanation:

`\sin( ( \alpha )/(2) ) = \sqrt{ (1 - \cos( \alpha ) )/(2) }`

`\cos( ( \alpha )/(2) ) = \sqrt{ (1 + \cos( \alpha ) )/(2) }`

Find cos using Pythagorean theorem.

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

`( (11)/(19) ) { }^(2) + ( \cos( \alpha ) ) {}^(2) = 1`

`( (121)/(281) ) + ( \cos( \alpha ) ) {}^(2) = 1`

`( \cos( \alpha ) ) {}^(2) = (160)/(281)`

`\cos( \alpha ) = (4 √(10) )/(19)`

Now, we use the formula

`\sin( ( \alpha )/(2) ) = \sqrt{ (1 - (4 √(10) )/(19) )/(2) }`

`\sin( ( \alpha )/(2) ) = \sqrt{ ( (19 - 4 √(10) )/(19) )/(2) }`

`\sin( ( \alpha )/(2) ) = \sqrt{ (19 - 4 √(10) )/(38) }`

`\cos( ( \alpha )/(2) ) = \sqrt{ (1 + \cos( \alpha ) )/(2) }`

`\cos( ( \alpha )/(2) ) = \sqrt{ (19 + 4 √(10) )/(38) }`