Question

If cos?=12÷13 find sine? and tan?

Answer 1

Given that \(\cos \theta = \frac{12}{13}\), we can find \(\sin \theta\) and \(\tan \theta\).

1. **Finding \(\sin \theta\)**:
Using the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]

Substituting \(\cos \theta = \frac{12}{13}\):
\[
\sin^2 \theta + \left(\frac{12}{13}\right)^2 = 1
\]
\[
\sin^2 \theta + \frac{144}{169} = 1
\]
\[
\sin^2 \theta = 1 - \frac{144}{169}
\]
\[
\sin^2 \theta = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}
\]
\[
\sin \theta = \sqrt{\frac{25}{169}} = \frac{5}{13}
\]

Since we don't have information about the quadrant in which \(\theta\) is located, both the positive and negative values are possible. Therefore, \(\sin \theta\) can be either \(\frac{5}{13}\) or \(-\frac{5}{13}\).

2. **Finding \(\tan \theta\)**:
Using the definition of tangent:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]

Substituting the values we found:
\[
\tan \theta = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}
\]

So, depending on the sign of \(\sin \theta\):
- If \(\sin \theta = \frac{5}{13}\): \(\tan \theta = \frac{5}{12}\)
- If \(\sin \theta = -\frac{5}{13}\): \(\tan \theta = -\frac{5}{12}\)

In conclusion:
- \(\sin \theta = \pm \frac{5}{13}\)
- \(\tan \theta = \frac{5}{12}\) or \(-\frac{5}{12}\) based on the sign of \(\sin \theta\).