Question

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

Answer 1

`2 / sqrt {2 - sqrt 2}`

Explanation:

`n = cossec \frac{7 pi}{8} = \frac{1}{sin \frac{7 pi}{8}} = \frac{1}{sin \frac{pi}{8}}`

`cos 2x = cos^2 x - sin^2 x = 1 - 2 sin^2 x`

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

`sqrt 2 /2 = 1 -2 frac{1}{n^2}`

`frac{2}{n^2} = 1 - sqrt 2 / 2`

`frac{2}{n^2} = (2 - sqrt 2) / 2`

`n^2 = 4 / (2 - sqrt 2)`

`n = 2 / sqrt {2 - sqrt 2}`

Answer 2

Find `csc ((7pi)/8)`

Ans: `2/(sqrt(2 - sqrt2)`

Explanation:

`csc ((7pi)/8) = 1 /sin ((7pi)/8)`
Find `sin ((7pi)/8)`. Call `sin ((7pi)/8) = sin t`
`cos 2t = cos ((14pi)/8) = cos ((7pi)/4) = cos (pi/4) = sqrt2/2`
Apply the trig identity: `cos 2t = 1 - 2sin^2 t`

`cos 2t = sqrt2/2 = 1 - 2sin^2 t`
`2sin^2 t = 1 - sqrt2/2 = (2 - sqrt2)/2`
`sin^2 t = (2 - sqrt2)/4`
`sin ((7pi)/8) = sin t = +- [sqrt(2 - sqrt2)]/2`
Since the arc (7pi/8) is located in Quadrant II. then only the positive answer is accepted.
`sin ((7pi)/8) = (sqrt(2 - sqrt2))/2`
Therefor, `csc ((7pi)/8) = (2)/(sqrt(2 - sqrt2)`