# given:tan=-2/3 and sin(theta)<0 find sin2(theta),cos2(theta) tan(theta) i need to know how to make the triangle. i have the triange x=3, y=-2. i get that part but how do i get r? r=square root of 13. i need the steps in between please. i dont understand how to get r Given `tan theta=-2/3` : The angle lies in the 2nd or 4th quadrants since the tangent is negative. Since we are given that `sintheta<0` we know it must be in the 4th quadrant.

So draw a triangle whose legs are parallel to the x and y axes, and the hypotenuse...

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Given `tan theta=-2/3` : The angle lies in the 2nd or 4th quadrants since the tangent is negative. Since we are given that `sintheta<0` we know it must be in the 4th quadrant.

So draw a triangle whose legs are parallel to the x and y axes, and the hypotenuse runs from the origin to the point (3,-2). The angle `theta` is formed by the positive x-axis and the hypotenuse.

The length of the hypotenuse can be found using the Pythagorean theorem: `r^2=3^2+(-2)^2==>r^2=13==>r=sqrt(13)`

(a) Now `sin2theta=2sinthetacostheta` . From the triangle we can see that `costheta=3/sqrt(13)` and `sintheta=-2/sqrt(13)` .

Then `sin2theta=2(3/sqrt(13))(-2/sqrt(13))=-12/13`

(b) Find `cos2theta tan theta` . `cos2theta=2cos^2theta-1` so:

`cos2thetatantheta=(2(3/sqrt(13))^2-1)(-2/3)=5/13(-2/3)=-10/39`

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