Question

point ( p ) lies on a unit circle at an angle of ( left(\frac{pi}{6}+pi
ight) ). what is the value of the ( y )-coordinate of point ( p )?
( \frac{sqrt{3}}{2} )
( \frac{1}{2} )
( -\frac{1}{2} )
( -\frac{sqrt{3}}{2} )

Explanation:

Step1: Recall the formula for the y - coordinate on a unit circle

On a unit circle, the \(y\) - coordinate of a point at an angle \(\theta\) is given by \(y = \sin\theta\). Here, \(\theta=\frac{\pi}{6}+\pi\).

Step2: Use the sine addition formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\)

We know that \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), where \(A=\frac{\pi}{6}\) and \(B = \pi\).
\(\sin(\frac{\pi}{6}+\pi)=\sin\frac{\pi}{6}\cos\pi+\cos\frac{\pi}{6}\sin\pi\)
Since \(\sin\frac{\pi}{6}=\frac{1}{2}\), \(\cos\pi=- 1\), \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\), and \(\sin\pi = 0\)
Substitute these values into the formula:
\(\sin(\frac{\pi}{6}+\pi)=\frac{1}{2}\times(-1)+\frac{\sqrt{3}}{2}\times0\)

Step3: Simplify the expression

\(\sin(\frac{\pi}{6}+\pi)=-\frac{1}{2}+0\)

Answer:

\(-\frac{1}{2}\) (the third option)