Question

Evaluate each of the following expressions by using special right triangles and the unit circle (give exact values),

Answer 1

Using equivalent and fundamenta. angles, we find that:

a) $\sin{\left(\frac{4\pi}{3}\right)} = \frac{\sqrt{3}}{2}$

b) $\cos{\left(\frac{11\pi}{3}\right)} = \frac{1}{2}$

c) $\sin{\left(-\frac{5\pi}{3}\right)} = -\frac{1}{2}$

d) $\cos{\left(-\frac{2\pi}{3}\right)} = -\frac{1}{2}$

Item a:

$\frac{4\pi}{3}$ is an angle in the third quadrant, as $\pi < \frac{4\pi}{3} < \frac{3\pi}{2}$

It's equivalent in the first quadrant is: $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$, which is a fundamental angle, which means that it's values of sine, cosine and tangent are known.

Sine in the third quadrant is negative, while in the first is positive, thus:

$\sin{\left(\frac{4\pi}{3}\right)} = -\sin{\left(\frac{\pi}{3}\right)} = \frac{\sqrt{3}}{2}$

Item b:

$\frac{11\pi}{3} = 2\pi + \frac{5\pi}{3}$

Thus, it is equivalent to an angle of $\frac{5\pi}{3}$, which is in the fourth quadrant.

It's equivalent on the first quadrant is:

$2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$

Cosine in the fourth quadrant is positive, just as in the first, thus:

$\cos{\left(\frac{11\pi}{3}\right)} = \cos{\left(\frac{\pi}{3}\right)} = \frac{1}{2}$

Item c:

$2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}$

Which is a angle in the third quadrant.

It's equivalent in the first is:

$\frac{7\pi}{6} - \pi = \frac{\pi}{6}$

Sine in the third quadrant is negative, while in the first is positive, thus:

$\sin{\left(-\frac{5\pi}{6}\right)} = \sin{\left(\frac{7\pi}{6}\right)} = -\sin{\left(\frac{\pi}{6}\right)} = -\frac{1}{2}$

Item d:

$2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$

Which is in the third quadrant.

Cosine in the third quadrant is negative, while in the first is positive, thus:

$\cos{\left(\frac{-2\pi}{3}\right)} = \cos{\left(\frac{4\pi}{3}\right)} = -\cos{\left(\frac{\pi}{3}\right)} = -\frac{1}{2}$

A similar problem is given at brainly.com/question/23843479