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Sagot :
Using equivalent and fundamenta. angles, we find that:
a) [tex]\sin{\left(\frac{4\pi}{3}\right)} = \frac{\sqrt{3}}{2}[/tex]
b) [tex]\cos{\left(\frac{11\pi}{3}\right)} = \frac{1}{2}[/tex]
c) [tex]\sin{\left(-\frac{5\pi}{3}\right)} = -\frac{1}{2}[/tex]
d) [tex]\cos{\left(-\frac{2\pi}{3}\right)} = -\frac{1}{2}[/tex]
Item a:
[tex]\frac{4\pi}{3}[/tex] is an angle in the third quadrant, as [tex]\pi < \frac{4\pi}{3} < \frac{3\pi}{2}[/tex]
It's equivalent in the first quadrant is: [tex]\frac{4\pi}{3} - \pi = \frac{\pi}{3}[/tex], 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:
[tex]\sin{\left(\frac{4\pi}{3}\right)} = -\sin{\left(\frac{\pi}{3}\right)} = \frac{\sqrt{3}}{2}[/tex]
Item b:
[tex]\frac{11\pi}{3} = 2\pi + \frac{5\pi}{3}[/tex]
Thus, it is equivalent to an angle of [tex]\frac{5\pi}{3}[/tex], which is in the fourth quadrant.
It's equivalent on the first quadrant is:
[tex]2\pi - \frac{5\pi}{3} = \frac{\pi}{3}[/tex]
Cosine in the fourth quadrant is positive, just as in the first, thus:
[tex]\cos{\left(\frac{11\pi}{3}\right)} = \cos{\left(\frac{\pi}{3}\right)} = \frac{1}{2}[/tex]
Item c:
[tex]2\pi - \frac{5\pi}{6} = \frac{7\pi}{6}[/tex]
Which is a angle in the third quadrant.
It's equivalent in the first is:
[tex]\frac{7\pi}{6} - \pi = \frac{\pi}{6}[/tex]
Sine in the third quadrant is negative, while in the first is positive, thus:
[tex]\sin{\left(-\frac{5\pi}{6}\right)} = \sin{\left(\frac{7\pi}{6}\right)} = -\sin{\left(\frac{\pi}{6}\right)} = -\frac{1}{2}[/tex]
Item d:
[tex]2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}[/tex]
Which is in the third quadrant.
Cosine in the third quadrant is negative, while in the first is positive, thus:
[tex]\cos{\left(\frac{-2\pi}{3}\right)} = \cos{\left(\frac{4\pi}{3}\right)} = -\cos{\left(\frac{\pi}{3}\right)} = -\frac{1}{2}[/tex]
A similar problem is given at https://brainly.com/question/23843479
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