Question

what is the exact value of \\(\cos(-60^\circ)\\)?
\\(\circ -\frac{\sqrt{3}}{2}\\)
\\(\circ -\frac{1}{2}\\)
\\(\circ \frac{1}{2}\\)
\\(\circ \frac{\sqrt{3}}{2}\\)

Explanation:

Step1: Recall the even function property of cosine

The cosine function is an even function, which means $\cos(-\theta) = \cos(\theta)$ for any angle $\theta$. So, we can rewrite $\cos(-60^\circ)$ as $\cos(60^\circ)$.

Step2: Find the value of $\cos(60^\circ)$

From the unit circle or special right triangles (30 - 60 - 90 triangle), we know that the cosine of $60^\circ$ is $\frac{1}{2}$. So, $\cos(-60^\circ)=\cos(60^\circ)=\frac{1}{2}$.

Answer:

$\frac{1}{2}$ (corresponding to the option "C. $\boldsymbol{\frac{1}{2}}$" if we assume the options are labeled A, B, C, D in order)