Logan Hill
09/16/2024 · High School
\( \operatorname{Cos} \theta=\frac{-24}{25} \) and \( 180^{\circ} \leqslant \theta \leqslant 270^{\circ} \)
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Step-by-step Solution
Given that \( \cos \theta = \frac{-24}{25} \) and the angle \( \theta \) is in the range \( 180^\circ \leq \theta \leq 270^\circ \), we can determine the sine of \( \theta \) using the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
First, we calculate \( \cos^2 \theta \):
\[
\cos^2 \theta = \left( \frac{-24}{25} \right)^2 = \frac{576}{625}
\]
Now, substituting this into the Pythagorean identity:
\[
\sin^2 \theta + \frac{576}{625} = 1
\]
To isolate \( \sin^2 \theta \), we subtract \( \frac{576}{625} \) from both sides:
\[
\sin^2 \theta = 1 - \frac{576}{625}
\]
Converting 1 to a fraction with a denominator of 625:
\[
1 = \frac{625}{625}
\]
Now we can perform the subtraction:
\[
\sin^2 \theta = \frac{625}{625} - \frac{576}{625} = \frac{49}{625}
\]
Taking the square root of both sides gives us:
\[
\sin \theta = \pm \sqrt{\frac{49}{625}} = \pm \frac{7}{25}
\]
Since \( \theta \) is in the third quadrant (where sine is negative), we have:
\[
\sin \theta = -\frac{7}{25}
\]
Thus, the values of \( \sin \theta \) and \( \cos \theta \) are:
\[
\sin \theta = -\frac{7}{25}, \quad \cos \theta = \frac{-24}{25}
\]
Quick Answer
\( \sin \theta = -\frac{7}{25} \), \( \cos \theta = \frac{-24}{25} \)
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