The terminal side of theta lies on the line 2x-y=0 in quadrant III, how do you find the values of the six trigonometric functions by finding a point on the line?

Feb 8, 2017

{:(sin(theta)=-2/sqrt(3),color(white)("XX"),csc(theta)=-sqrt(3)/2), (cos(theta)=-1/sqrt(3),,sec(theta)=-sqrt(3)), (tan(theta)=2,,cot(theta)=1/2) :}

Explanation:

Any relation of the form $A x + B y = C$ has a slope of (i.e. a $\tan$) $- \frac{A}{B}$
So $2 x - y = 0$ has a slope ($\tan$) of $2$.

Furthermore, we can see that $2 x - y = 0$ passes through the origin,
so we have the situation in the image below for a point on the line in Q III: The six trigonometric values can be determined by applying their definitions based on "adjacent", "opposite", and "hypotenuse" sides.

Apr 12, 2017

$\sin \theta = - \frac{2 \sqrt{5}}{5}$
$\cos \theta = - \frac{\sqrt{5}}{5}$
Because $\sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$
EDIT: How I did this was by picking a relivant point that I could work with. You could also use the unit circle equation of ${x}^{2} + {y}^{2} = 1$ to find the answer as well. With 2x - y = 0 moving the y to the right you end up with 2x = y (y = 2x). your unit circle equation becomes ${x}^{2} + {\left(2 x\right)}^{2} = 1$ and you solve from there to end up with the answers above.