Question

Find the exact values of six trigonometric functions of the angle theta that has a terminal angle at point (-2,6)

Answer 1

A figure will be very helpful.

Answer 2

I do not have a figure to give you

Answer 3

Given (x, y) = (-2, 6) ----point lies in the 2nd quadrant..
Let r represents the length of the hypotenuse, therefore..
\[\huge r^2 = x^2+y^2 = 4+ 36 = 40 \rightarrow r =2 \sqrt{10}\]
Let the angle be \[\huge \theta\]
Hence,
\[\huge \sin \theta = \frac{y}{r} = \frac{6}{2 \sqrt {10}}= \frac{3}{ \sqrt {10}}\]

\[\huge \cos \theta = \frac{x}{r} = \frac{-2}{2 \sqrt {10}}= -\frac{1}{ \sqrt {10}}\]

\[\huge \tan \theta = \frac{y}{x} = \frac{-2}{6}= -\frac{1}{3} \]
\[\huge \cot \theta = \frac{1}{\tan \theta} =\frac{-3}{1}=-3 \]

\[\huge \sec \theta = \frac{1}{\cos \theta} =- \frac{\sqrt{10}}{1}=-\sqrt{10} \]

\[\huge cosec \theta = \frac{1}{\sin \theta} =\frac{\sqrt{10}}{3} \]

@sierra24