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

1 Answer
Aug 1, 2016

Answer:

#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#