Question

How do you find the exact values of `sin2u, cos2u, tan2u` using the double angle values given `cscu=3, pi/2<u<pi`?

Answer 1

`csc u =1/(sin u) = 3` --> `sin u = 1/3`.
Find cos u
`cos^2 u = 1 - sin^2 u = 1 - 1/9 = 8/9`
`cos u = +- (2sqrt2)/3`
Since u is in Quadrant 2, cos u is negative.
`sin 2u = 2sin u.cos u = 2(1/3)(- 3/(2sqrt3)) = - 1/sqrt3 = - sqrt3/3`
`cos 2u = 1 - 2sin^2 u = 1 - 2/9 = 7/9`
`tan 2u = sin/(cos) = (- sqrt3/3)(9/7) = - (3sqrt3)/7`