Question

How do you use the half-angle identity to find the exact value of sin67.5°?

Answer 1

Find `sin 67^@5`

Ans: sqrt(2 + sqrt2)/2

Explanation:

Call sin (67.5) = sin t
cos 2t = cos 135 = -sqrt2/2 (Trig Table of Special Arcs).
Apply the trig identity: `cos 2t = 1 - 2sin^2 t.`
`-sqrt2/2 = 1 - 2sin^2 t`
`2sin^2 t = 1 + sqrt2 = (2+sqrt2)/2`
`sin^2 t = (2 + sqrt2)/4`
`sin t = sqrt(2 + sqrt2)/2`. Only the positive answer is accepted since the arc 67.5 is in Quadrant I.
`sin t = sin (67.5) = sqrt(2 + sqrt2)/2`.

Check by calculator>
sin (67.5) = 0.92
`sqrt(2 + sqrt2)/2 = 1.85/2 = 0. 92`. OK