Solve for the exact values of sine and cosine for the angle $frac{7pi}{6}$ on the unit circle.
Answer 1
Thomas Walker
First, note that $\frac{7\pi}{6}$ radians is in the third quadrant where both sine and cosine are negative. The reference angle for $\frac{7\pi}{6}$ is $\pi – \frac{7\pi}{6} = \frac{\pi}{6}$.
The sine and cosine values for $\frac{\pi}{6}$ are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, respectively. Since $\frac{7\pi}{6}$ is in the third quadrant, these values become negative.
Therefore, the exact values are:
$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$
$\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
Answer 2
Alex Thompson
To solve for the sine and cosine of $frac{7pi}{6}$, recognize it falls in the third quadrant where sine and cosine are negative.
The reference angle is $pi – frac{7pi}{6} = frac{pi}{6}$, where the sine and cosine values for $frac{pi}{6}$ are $frac{1}{2}$ and $frac{sqrt{3}}{2}$, respectively.
Thus:
$sinleft(frac{7pi}{6}
ight) = -frac{1}{2}$
$cosleft(frac{7pi}{6}
ight) = -frac{sqrt{3}}{2}$
Answer 3
William King
Find the reference angle for $frac{7pi}{6}$, which is $frac{pi}{6}$. In the third quadrant, both sine and cosine are negative:
$sinleft(frac{7pi}{6}
ight) = -frac{1}{2}$
$cosleft(frac{7pi}{6}
ight) = -frac{sqrt{3}}{2}$
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