Question

If tanx= -1/3, cos>0, then how do you find sin2x?

Answer 1

`sin(2x) = -0.6`

Explanation:

Any trigonometric function of some angle can be easily expressed in terms of a tangent of half of this angle.

We can express `sin(2x)` in terms of `tan(x)` as follows:
`sin(2x) = 2sin(x)cos(x) = 2sin(x)/cos(x)*cos^2(x)=2tan(x)*cos^2(x)`

In its turn,
`1/cos^2(x) = [sin^2(x)+cos^2(x)]/cos^2(x) = 1+sin^2(x)/cos^2(x) = 1+tan^2(x)`

Therefore,
`sin(2x) = (2tan(x))/(1+tan^2(x))`

Using this and given value `tan(x) = -1/3`, we conclude
`sin(2x) = 2(-1/3)/(1+1/9) =-6/10=-0.6`