Question

If cos?=12÷13 find sine? and cos?

Answer 1

To find \(\sin \theta\) and \(\cos \theta\) given that \(\cos \theta = \frac{12}{13}\), we can use the Pythagorean identity:

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

We know \(\cos \theta = \frac{12}{13}\), so we can calculate \(\cos^2 \theta\):

\[
\cos^2 \theta = \left(\frac{12}{13}\right)^2 = \frac{144}{169}
\]

Now, we substitute this into the Pythagorean identity:

\[
\sin^2 \theta + \frac{144}{169} = 1
\]

To isolate \(\sin^2 \theta\), subtract \(\frac{144}{169}\) from both sides:

\[
\sin^2 \theta = 1 - \frac{144}{169}
\]

To combine the terms, note that \(1\) can be rewritten as \(\frac{169}{169}\):

\[
\sin^2 \theta = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}
\]

Now, take the square root of both sides to find \(\sin \theta\):

\[
\sin \theta = \sqrt{\frac{25}{169}} = \frac{5}{13}
\]

Since the value of \(\theta\) is not specified, \(\sin \theta\) can be either positive or negative. Typically, if we consider \(\theta\) in the first quadrant (where \(\sin\) and \(\cos\) are both positive), we take \(\sin \theta = \frac{5}{13}\).

Thus, we have:

\[
\sin \theta = \frac{5}{13}, \quad \cos \theta = \frac{12}{13}
\]