Question

Given that `sin(b)=-2/3`, and that `b` is in the third quadrant, what is the value of `tan(2b)`..?

Answer 1

Given `sinb=-2/3," where "b in "III quadrant"`

So `cosb=-sqrt(1-sin^2b)=sqrt(1-4/9)=-sqrt5/3`

Now `tan(2b)=(sin2b)/(cos2b)`

`=(2sinbcosb)/(cos^2b-sin^2b)`

`=(2(-2/3)(-sqrt5/3))/(5/9-4/9)`

`=4sqrt5`