Trigonometry / Finding the exact value

Question

Given that $\cos \theta = \dfrac{-4}{5}$ and $\sin \theta$ is positive, obtain the exact values of

$\cos (6\pi+\theta)$

i don't understand this question.

Answer 1

Notice that $\cos(\theta)$ is periodic with period $2\pi$, meaning that $$\cos(\theta) \equiv \cos(\theta+2\pi)\equiv\cos(\theta+4\pi) \equiv \cos(\theta+6\pi)\equiv \cos(\theta+2n\pi)$$ for $n \in \mathbb{Z}.$

Using this fact, we see that, in this case, $$\boxed{\cos(\theta+6\pi)=\cos(\theta)=-\frac{4}{5}\ \ }.$$

Answer 2

If you use that $$\cos(a+b)=\cos a \cos b-\sin a \sin b,$$ then you get $\cos (6\pi+\theta)=\cos \theta=-4/5$.