Answer:
-0.555
Explanation:
The terminal point of the vector in this problem is
(-2,-3)
So, it is in the 3rd quadrant.
We want to find the angle
that gives the direction of this vector.
We can write the components of the vector along the x- and y- direction as:
![v_x = -2\\v_y = -3](https://img.qammunity.org/2021/formulas/mathematics/high-school/jice62e6ezi353j9i41c8l6nltqf77cy8j.png)
The tangent of the angle will be equal to the ratio between the y-component and the x-component, so:
![tan \theta = (v_y)/(v_x)=(-3)/(-2)=1.5\\\theta=tan^(-1)(1.5)=56.3^(\circ)](https://img.qammunity.org/2021/formulas/mathematics/high-school/peuo0vg9p322o7o6wbl9pk9ojtcji2amz2.png)
However, since we are in the 3rd quadrant, the actual angle is:
![\theta=180^(\circ) + 56.3^(\circ) = 236.3^(\circ)](https://img.qammunity.org/2021/formulas/mathematics/high-school/d70gtj8e4asnjrz1h4wgxhriiiyg98scfw.png)
So now we can find the cosine of the angle, which will be negative:
![cos \theta = cos(236.3^(\circ))=-0.555](https://img.qammunity.org/2021/formulas/mathematics/high-school/bqsg287g986znzyh0za3vurcac5bnvyh4n.png)