Question

How do you find the value for `sin2theta`, `cos2theta`, and `tan2theta` and the quadrant in which `2theta` lies given `sintheta=-sqrt10/10` and `theta` is in quadrant IV?

Answer 1

1t is in Quadrant 4.
`sin 2t = - 3/5`
`cos 2t = 4/5`
`tan 2t = - 3/4`

Explanation:

`sin t = - 1/sqrt10`. Find cos t
`cos^2 t = 1 - sin^2 t = 1 - 1/10 = 9/10`
`cos t = 3/sqrt10` --> because t is in Quadrant 4
`sin 2t = 2sin t.cos t = 2(-1/sqrt10)(3/sqrt10) = - 6/10= - 3/5`
Find cos 2t.
`cos 2t = 2cos^2 t - 1 = 18/10 - 1 = 8/10 = 4/5`
2t is in Quadrant 4 because its sin is negative and its cos is positive.
`tan 2t = (sin 2t)/(cos 2t) = (-3/5)(5/4) = - 3/4`