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Consider this equation.
[tex]\[ \sin (\theta) = -\frac{4 \sqrt{29}}{29} \][/tex]

If [tex]\(\theta\)[/tex] is an angle in quadrant IV, what is the value of [tex]\(\cos (\theta)\)[/tex]?

A. [tex]\(-\frac{\sqrt{377}}{29}\)[/tex]
B. [tex]\(\frac{\sqrt{377}}{29}\)[/tex]
C. [tex]\(\frac{4 \sqrt{29}}{29}\)[/tex]
D. [tex]\(-\frac{4 \sqrt{29}}{29}\)[/tex]


Sagot :

We are given:
[tex]\[ \sin (\theta) = -\frac{4 \sqrt{29}}{29} \][/tex]
and we need to find the value of [tex]\(\cos (\theta)\)[/tex] given that [tex]\(\theta\)[/tex] is in quadrant IV.

First, let's recall that in quadrant IV, the sine function is negative, and the cosine function is positive. This is consistent with our given value of [tex]\(\sin(\theta)\)[/tex].

Next, we use the Pythagorean identity, which states:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]

Substitute the given [tex]\(\sin (\theta)\)[/tex]:
[tex]\[ \sin(\theta) = -\frac{4 \sqrt{29}}{29} \][/tex]

Calculate [tex]\(\sin^2 (\theta)\)[/tex]:
[tex]\[ \sin^2(\theta) = \left(-\frac{4 \sqrt{29}}{29}\right)^2 = \frac{(4 \sqrt{29})^2}{29^2} = \frac{16 \cdot 29}{841} = \frac{464}{841} \][/tex]

Next, we use the Pythagorean identity to find [tex]\(\cos^2 (\theta)\)[/tex]:
[tex]\[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{464}{841} = \frac{841}{841} - \frac{464}{841} = \frac{377}{841} \][/tex]

Now, take the square root of both sides to find [tex]\(\cos (\theta)\)[/tex]. Since [tex]\(\theta\)[/tex] is in quadrant IV, [tex]\(\cos (\theta)\)[/tex] is positive:
[tex]\[ \cos(\theta) = \sqrt{\frac{377}{841}} = \frac{\sqrt{377}}{29} \][/tex]

The correct answer is:
B. [tex]\(\frac{\sqrt{377}}{29}\)[/tex]
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