2 trig questions?

Question

i'm really bad in graphing. the question is:

find sin θ and tan θ if the terminal side of θ lies along the line y = 2x in quadrant III. my question is how exacty do u find the (x, y) for y = 2x?

and a similar question: find sin θ and tan θ if the terminal side of θ lies along the line y = -3x in quadrant II. How do u find the (x,y) for y = -3x?

Answer 1

Did you try graphing y=2x? :) You can pick any x and y where the y is twice as big as the x...like (1,2) or (2,4). The point doesn't matter. The ratio is what matters...and the hypotenuse.

Suppose you pick (1,2). Go ahead and draw it. You'll see that, if you drop a line straight down to the x-axis, you've got a right triangle. So the length of the hypotenuse is sqrt(1^2 + 2^2) = sqrt(5). So now you've got the opposite (2), the adjacent (1), and the hypotenuse (sqrt(5))... you can take it from there right?

Answer 2

I can't draw a sketch here, but you can do it on paper.

Line y = 2x has gradient 2, so if you go across 1 unit left from (0,0) you have to go 2 units down to get to the line. The terminal side so formed has length sqrt(5). You can divide by sqrt(5) to make a triangle with terminal side 1 unit long. Then the (x, y) values also have to be divided by sqrt(5) and you have the coordinates of the end of the terminal side as
(-1/sqrt(5), -2/sqrt(5)).
By definition, sin θ is the y value and cos θ is the x value.
tan θ is the gradient of the line.

Do similar work for the second one, but go 1 unit left and 3 up, then divide by sqrt(10)

Answer 3

Use any two arbitrary values for x, and calculate y for each. In this case, very convenient if you use 0 and 1.

For y=2x, you get the points (0.0) and (-1,-2) for two points lying in the III quadrant.

For y=-3x, you have (0,0) and (-1, 3).

Answer 4

answer to both questions: it doesn't matter what that actuall values of x and y are: pick arbitrary values that fit the relationship and plug 'em in:

try (1, 2) and (2, 4) for the first question: the answer will be the same

try (1, -3) and (2, -6) for the second: the answer will be the same

-it's the ratio of x to y that matters.