Question

How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

Answer 1

`sin (t/2) = sqrt10/10`
`cos (t/2) = (3sqrt10)/10`

Explanation:

`cos t = 4/5`. Find `sin (t/2) and cos (t/2)`.
Use the 2 trig identities:
`2cos ^2 (t/2) = 1 + cos t`
`2sin^2 (t/2) = 1 - cos t`
We have:
`2cos^2 (t/2) = 1 + cos t = 1 + 4/5 = 9/5`
`cos^2 (t/2) = 9/10`
`cos (t/2) = +- 3/sqrt10 = +- (3sqrt10)/10`
Since t is in Quadrant I, `cos (t/2)` is positive --> `cos t/2 = (3sqrt10)/10`
`2sin^2 (t/2) = 1 - cos t = 1 - 4/5 = 1/5`
`sin^2 (t/2) = 1/10`
`sin (t/2) = +- 1/sqrt10 = +- sqrt10/10`
Since t is in quadrant I, `sin (t/2)` is positive -->
`sin (t/2) = sqrt10/10`