Question

If `sinu=-4/5` and `pi < u < (3pi)/2`, find `sin2u,cos2u` and `tan2u`?

Answer 1

`sin2u=24/25`, `cos2u=-7/25` and `tan2u=-24/7`

Explanation:

As `pi < u < (3pi)/2`, `u` is in `Q3` and `cosu<0`.

As `sinu=-4/5`, `cosu=-sqrt(1-(-4/5)^2)=-3/5`

Hence, `sin2u=2sinucosu=2xx(-4/5)xx(-3/5)=24/25`

`cos2u=cos^2u-sin^2u=9/25-16/25=-7/25`

and `tan2u=(sin2u)/(cos2u)=(24/25)/(-7/25)=-24/7`