Question

which of the following best explains the value of sin(π/3) on the unit circle below? (0, 1) π/2 r = 1 π/3 0 (-1, 0) π (1, 0) 3π/2 (0, -1) sin(π/3) = opposite/hypotenuse = opposite/1 = opposite

Explanation:

Step1: Recall sine - definition on unit circle

On a unit circle with radius $r = 1$, for an angle $\theta$, $\sin\theta=\frac{y}{r}$. Since $r = 1$, $\sin\theta=y$. For the angle $\theta=\frac{\pi}{3}$, we consider the right - triangle formed in the unit circle. The sine of an angle in a right - triangle is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. In the unit circle, the hypotenuse is the radius $r = 1$.

Step2: Determine the value of $\sin\frac{\pi}{3}$

For $\theta=\frac{\pi}{3}$, the coordinates of the point on the unit circle are $(\frac{1}{2},\frac{\sqrt{3}}{2})$. Since $\sin\theta$ is the $y$ - coordinate of the point on the unit circle corresponding to the angle $\theta$, $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$. And from the right - triangle perspective with hypotenuse $r = 1$, $\sin\frac{\pi}{3}=\frac{\text{opposite}}{1}=\text{opposite}$, where the length of the opposite side to the angle $\frac{\pi}{3}$ in the right - triangle formed in the unit circle is $\frac{\sqrt{3}}{2}$.

Answer:

$\sin\frac{\pi}{3}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{opposite}}{1}=\text{opposite}$, and the value of the opposite side (the $y$ - coordinate of the point on the unit circle corresponding to the angle $\frac{\pi}{3}$) is $\frac{\sqrt{3}}{2}$