Question

If cos 0= 1/4 and 0 is in quadrant I find sin 0

Answer 1

To find \(\sin \theta\) given that \(\cos \theta = \frac{1}{4}\) and \(\theta\) is in the first quadrant, we can use the Pythagorean identity:

\[
\sin^2 \theta + \cos^2 \theta = 1
\]

Substituting \(\cos \theta\) into the equation, we have:

\[
\sin^2 \theta + \left(\frac{1}{4}\right)^2 = 1
\]

Calculating \(\left(\frac{1}{4}\right)^2\):

\[
\sin^2 \theta + \frac{1}{16} = 1
\]

Subtracting \(\frac{1}{16}\) from both sides gives:

\[
\sin^2 \theta = 1 - \frac{1}{16}
\]

Converting 1 into a fraction with a denominator of 16:

\[
1 = \frac{16}{16}
\]

So we have:

\[
\sin^2 \theta = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}
\]

Now, we 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 the first quadrant, \(\sin \theta\) is positive. Therefore, the final answer is:

\[
\sin \theta = \frac{\sqrt{15}}{4}
\]