Question

David is told that an angle has a cosine of - sqrt(3)/2 and is in Quadrant III. Use the Pythagorean identity sin^2(θ )+cos^2(θ )=1 and the quadrant to find the sine.

Answer 1

To find the sine of an angle \( \theta \) given that \( \cos(\theta) = -\frac{\sqrt{3}}{2} \) and that \( \theta \) is in Quadrant III, we can use the Pythagorean identity:

\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

First, we substitute the value of \( \cos(\theta) \):

\[
\sin^2(\theta) + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1
\]

Calculating \( \left(-\frac{\sqrt{3}}{2}\right)^2 \):

\[
\left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}
\]

Now substitute this back into the equation:

\[
\sin^2(\theta) + \frac{3}{4} = 1
\]

Next, we isolate \( \sin^2(\theta) \):

\[
\sin^2(\theta) = 1 - \frac{3}{4}
\]

This simplifies to:

\[
\sin^2(\theta) = \frac{1}{4}
\]

Now we take the square root of both sides:

\[
\sin(\theta) = \pm \frac{1}{2}
\]

Since \( \theta \) is in Quadrant III, where sine is negative, we choose the negative root:

\[
\sin(\theta) = -\frac{1}{2}
\]

Thus, the sine of the angle \( \theta \) is:

\[
\sin(\theta) = -\frac{1}{2}
\]