# 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?

Any relation of the form

So

Furthermore, we can see that

so we have the situation in the image below for a point on the line in Q III:

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The answer is:

Because

Using the distance formula you arrive at the square root of 5. This answer is verified via a Larson Precalculus book in the answer section.

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To find the values of the six trigonometric functions for an angle ( \theta ) whose terminal side lies on the line ( 2x - y = 0 ) in quadrant III, we need a point on the line that also lies in quadrant III, where both ( x ) and ( y ) are negative.

The equation ( 2x - y = 0 ) can be rewritten as ( y = 2x ). We need a point ((x, y)) on this line in the third quadrant. A simple choice would be ( x = -1 ), which gives ( y = -2 ) (since ( y = 2(-1) = -2 )).

So, we have the point ((-1, -2)) on the line in quadrant III. This point means that if you draw a right triangle from the origin to this point, the lengths of the sides relative to the origin would be:

- ( x = -1 ) (the horizontal distance, or the adjacent side of the angle ( \theta )),
- ( y = -2 ) (the vertical distance, or the opposite side of the angle ( \theta )),
- The hypotenuse ( r ) can be found using the Pythagorean theorem: ( r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} ).

The six trigonometric functions of ( \theta ) are defined as follows:

- Sine (( \sin(\theta) )) is opposite/hypotenuse: ( \sin(\theta) = \frac{-2}{\sqrt{5}} ).
- Cosine (( \cos(\theta) )) is adjacent/hypotenuse: ( \cos(\theta) = \frac{-1}{\sqrt{5}} ).
- Tangent (( \tan(\theta) )) is opposite/adjacent: ( \tan(\theta) = \frac{-2}{-1} = 2 ).
- Cosecant (( \csc(\theta) )) is the reciprocal of sine: ( \csc(\theta) = \frac{\sqrt{5}}{-2} ).
- Secant (( \sec(\theta) )) is the reciprocal of cosine: ( \sec(\theta) = \frac{\sqrt{5}}{-1} ).
- Cotangent (( \cot(\theta) )) is the reciprocal of tangent: ( \cot(\theta) = \frac{1}{2} ).

These values are based on the point ((-1, -2)) and respect the signs of trigonometric functions in quadrant III, where both sine and cosine are negative, and consequently, their reciprocals are also negative.

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