1 multiple choice 1 point evaluate the trigonometric function of the given quadrantal angle. cos 13π 1 -1 0…

Explanation:

Step1: Use the periodicity of cosine function

The cosine function has a period of (2\pi), i.e., (\cos(x + 2k\pi)=\cos x) for any integer (k). We can write (13\pi) as (12\pi+\pi), so (\cos(13\pi)=\cos(12\pi + \pi)). Since (12\pi = 6\times2\pi), then (\cos(12\pi+\pi)=\cos\pi) (using (\cos(x + 2k\pi)=\cos x) with (x = \pi) and (k = 6)).

Step2: Evaluate (\cos\pi)

We know that on the unit - circle, for the angle (\theta=\pi), the coordinates of the point on the unit - circle are ((- 1,0)). By the definition of the cosine function (\cos\theta=x) (where ((x,y)) is the point on the unit - circle corresponding to the angle (\theta)), when (\theta=\pi), (x=-1). So (\cos\pi=-1).

Answer:

(-1) (corresponding to the second option)