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
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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)` .
(b) Find `cos2theta tan theta` . `cos2theta=2cos^2theta-1` so:
To find the length of r, you will want to use the Pythagorean Theorem a^2+b^2=c^2, where a=3, b=-2, and c=r. Then solve for your unknown length r.
From there, use trig identities to solve for sin(theta) and cos(theta).
sin(theta)=y/r = -2/sqroot(13)
cos(theta)=x/r = 3/sqroot(13)
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