To find \( \sin \theta \) given \( \cos \theta = \frac{1}{4} \) and that \( \theta \) is in Quadrant I, we can use the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
First, we know that:
\[ \cos^2 \theta = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Now, substitute \( \cos^2 \theta \) into the Pythagorean identity:
\[ \sin^2 \theta + \frac{1}{16} = 1 \]
To isolate \( \sin^2 \theta \), subtract \( \frac{1}{16} \) from both sides:
\[ \sin^2 \theta = 1 - \frac{1}{16} \]
Convert 1 to have a denominator of 16:
\[ 1 = \frac{16}{16} \]
Now, perform the subtraction:
\[ \sin^2 \theta = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \]
Next, take the square root of both sides to find \( \sin \theta \):
\[ \sin \theta = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]
Since \( \theta \) is in Quadrant I, where sine is positive:
\[ \sin \theta = \frac{\sqrt{15}}{4} \]
Thus, the final answer is:
\[ \sin \theta = \frac{\sqrt{15}}{4} \]
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