Question

How do you find the exact values of `sin2u, cos2u, tan2u` using the double angle values given `secu=-5/2, pi/2<u<pi`?

Answer 1

`sin 2u = - (4sqrt21)/25`
`cos 2u = - 17/25`
`tan 2u = (4sqrt21)/17`

Explanation:

`sec u = 1/(cos u) = - 5/2` --> `cos u = - 2/5`.
Find `sin u`
`sin ^2 u = 1 - cos^2 u = 1 - 4/25 = 21/25`
`sin u = +- sqrt21/5`.
Since `u` is in Quadrant II, then `sinu` is positive.

`sin2u=2sinucosu=2(sqrt21/5)(-2/5) =- (4sqrt21)/25`

`cos 2u = 2cos^2 u - 1 = 2(4/25) - 1 = - 17/25`
`tan 2u = sinu/cosu = (- (4sqrt21)/25)(- 25/17) = (4sqrt21)/17`