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german
2 years ago
5

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

Mathematics
1 answer:
levacccp [35]2 years ago
3 0

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

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