Question

if $\theta=\frac{-1\pi}{6}$, then $\sin(\theta)=\square$ $\cos(\theta)=\square$ give exact values. no decimals allowed! example: enter sqrt(2)/2 for $\frac{\sqrt{2}}{2}$. with functions like sqrt, be sure to use function notation (parentheses). sqrt(2)/2 will work, but sqrt2/2 will not. question help: video 1 video 2

Explanation:

Step1: Use the property of sine function

Since \(\sin(-\alpha)=-\sin\alpha\), for \(\theta =-\frac{\pi}{6}\), we have \(\sin(-\frac{\pi}{6})=-\sin\frac{\pi}{6}\).
And we know that \(\sin\frac{\pi}{6}=\frac{1}{2}\), so \(\sin(-\frac{\pi}{6})=-\frac{1}{2}\).

Step2: Use the property of cosine function

Since \(\cos(-\alpha)=\cos\alpha\), for \(\theta =-\frac{\pi}{6}\), we have \(\cos(-\frac{\pi}{6})=\cos\frac{\pi}{6}\).
And we know that \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\).

Answer:

\(\sin(\theta)=-\frac{1}{2}\), \(\cos(\theta)=\frac{\sqrt{3}}{2}\)