Question

How do you use half-angle identities to find the exact values of sin (u/2) & cos (u/2) when sin(u)= - 4/5 and 3pi/2 <u<2pi?

Answer 1

`cos (u/2) = -2sqrt5/5`
`sin (u/2) = sqrt5/5`

Explanation:

`sin u = - 4/5` --> `cos ^2 u = 1 - sin^2 u = 1 - 16/25 = 9/25`
`cos u = 3/5` (cos u positive since u is in the Quadrant IV).
To find `sin (u/2)` and `cos (u/2)` apply the trig identities:
`cos 2a = 2cos^2 a - 1`
`cos 2a = 1 - 2sin^2 a`
a. Find `cos (u/2)`
`cos u = 3/5 = 2cos^2 (u/2) - 1`
`2cos^2 (u/2) = 1 + 3/5 = 8/5`
`cos^2 (u/2) = 8/10`
`cos (u/2) = sqrt8/sqrt10 = +- sqrt80/10 = +- 4sqrt5/10 = +- 2sqrt5/5`
`cos (u/2) = - 2sqrt5/5` (`cos (u/2)` negative, Quadrant II)
b. Find `sin (u/2)`
`3/5 = 1 - 2sin^2 (u/2)`
`2sin^2 (u/2) = 1 - 3/5 = 2/5`
`sin^2 (u/2) = 2/10 = 1/5`
`sin (u/2) = +- 1/sqrt5 = +- sqrt5/5`
`sin (u/2) = sqrt5/5` (`sin (u/2)` positive, Quadrant II)