How do you find the remaining trigonometric functions of theta given sin theta=12/13 and theta terminates in QI?

Mar 1, 2017

$\sin x = \frac{12}{13}$, $\cos x = \frac{5}{13}$, $\tan x = \frac{12}{5}$

$\cot x = \frac{5}{12}$, $\sec x = \frac{13}{5}$ and $\csc x = \frac{13}{12}$

Explanation:

As $\theta$ terminates in $Q I$, all trigonometric ratios are positive.

$\therefore \cos x = \sqrt{1 - {\sin}^{2} x} = \sqrt{1 - {\left(\frac{12}{13}\right)}^{2}}$

= $\sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$

and $\tan x = \sin \frac{x}{\cos} x = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{13} \times \frac{13}{5} = \frac{12}{5}$

and $\cot x = \frac{1}{\tan} x = \frac{5}{12}$, $\sec x = \frac{1}{\cos} x = \frac{13}{5}$ and

$\csc x = \frac{1}{\sin} x = \frac{13}{12}$