Find the exact value of $\sin ( 2 \pi )$ and $\cos ( - 2 \pi )$
A) $\sin ( 2 \pi ) = 1 , \quad \cos ( - 2 \pi ) = - 1$
B) $\sin ( 2 \pi ) = 0 , \quad \cos ( - 2 \pi ) = - 1$
C) $\sin ( 2 \pi ) = 1 , \quad \cos ( - 2 \pi ) = 0$
D) $\sin ( 2\pi ) = 0 , \quad \cos ( - 2 \pi ) = - 1$
E) none of these

Question

Quizplus · Accepted Answer

$\sin(2\pi) = 0$ because $2\pi$ radians corresponds to a full circle, bringing the sine function back to its starting point at 0. $\cos(-2\pi) = 1$ because cosine is an even function, meaning $\cos(-\theta) = \cos(\theta)$, and $\cos(2\pi) = 1$ at a full circle. None of the provided options correctly match both values.

Find the Exact Value Of sin⁡(2π)\sin ( 2 \pi )sin(2π) And cos⁡(−2π)\cos ( - 2 \pi )cos(−2π)

Find the Exact Value Of $\sin ( 2 \pi )$ And $\cos ( - 2 \pi )$