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?

Question

Delilah Aguilar · Accepted Answer

#{:(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) :}#

Any relation of the form #Ax+By=C# has a slope of (i.e. a #tan#) #-A/B#
So #2x-y=0# has a slope (#tan#) of #2#.

Furthermore, we can see that #2x-y=0# passes through the origin,
so we have the situation in the image below for a point on the line in Q III:

Anna Andrews · Answer

The answer is:
#sin theta = -(2sqrt(5))/5#

#cos theta = - sqrt(5)/5#

Because #sqrt(2^2 + (-1)^2) = sqrt(5)# 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. 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 + (2x)^2 = 1# and you solve from there to end up with the answers above.

Brooklyn Anderson · Answer

undefined 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.