# angle theta in standard position

I have no idea how to start this homework. Here is the question.

Sketch a angle theta in standard position such that theta has the least possible positive measure, and the given point is on the terminal side of theta. Then find the values of the six trigonometric functions for each angle. Rationalize denominators when possible.

One of the questions is $(5, -12)$ I have no idea what is going on. I mean -1,000,000,000(theta) would be very negative, but not the most negative, so what do I do?

Also the next set it just as confusing. Suppose that the point x,y is in the indicated quadrants. Decide whether the given ratio is pos or neg. Recall that $r = \sqrt{x^2 +y^2}$ hint drawing a sketch may help.

II x/r

What do I do? I know y is positive and x negative so that makes it negative correct? One more quick question I have the same problem except I am given a line $2x+y=0$ $x\ge 0$. What points on the ling do I take since it goes on to infinity?

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"I am given a line ... ". Hint: note that equation $2x+y=0$ is equivalent to $y=-2x$; and $r=\sqrt{x^2+y^2}=\sqrt{x^2+(-2x)^2}=\sqrt{x^{2}+4x^{2}}=\sqrt{5x^{2}}=\sqrt{5}\‌​left\vert x\right\vert$ –  Américo Tavares Jun 1 '11 at 19:48
I don't get what that does at all, sorry. I think I did get the answer by making x=2 and y=-4 –  Adam Jun 1 '11 at 20:03
I assume you are given the line $2x+y=0$ for $x\ge 0$ and are asked to find whether $x/r$ is positive or negative. My hint was intended to show that if $x$ is positive, so is $x/r$, because $r\ge 0$. Your choice of $x=2$ and $y=-4$ is enough and simpler for that purpose. –  Américo Tavares Jun 1 '11 at 20:25

All of the fancy language unfortunately muddies what is a simple situation. Draw axes as usual, and identify the point $P=(5, -12)$. Let $\theta$ be the smallest positive angle through which you need to rotate the positive $x$-axis so that it will pass through $P$.

Remember that rotation counterclockwise is positive by definition. One says "the smallest angle" (or in fancy language, the angle of smallest positive measure) because if rotating counterclockwise through $\theta$ gets you to $P$, then so does rotation through $\theta +2\pi$ (in radians) or rotation through $\theta +360^\circ$, if you are using degrees. Also, rotation through $\theta+4\pi$ would get you to the same place, and so on.

Now it should not be hard to find $\sin\theta$, $\cos\theta$, and so on. If you wish, I will amplify the post to deal with them all. The point $P$ is at distance $$r=\sqrt{(5)^2+(-12)^2}=13$$ from the origin. Now $\cos\theta=x/r=5/13$, $\sin\theta=y/r=-12/13$, and the other trigonometric functions are then easy to calculate.

Added: To complete the list, $\tan\theta=\sin\theta/\cos\theta=y/x$, $\cot\theta=1/\tan\theta=x/y$, $\sec\theta=1/\cos\theta=r/x$, $\csc\theta=1/\sin\theta=r/y$. For all of these functions, if the denominator is $0$, we say that the function does not exist (or is not defined) for the relevant angle. For instance, the cosine of a right angle is $0$, so $\sec(\pi/2)$ does not exist.

Added: About the new "$2x+y$" problem, you do not say what the problem is. I am guessing you are asked something similar to the $(5,-12)$ problem. If so, take any positive $x$, preferably simple like $x=1$, but it doesn't matter. Then $y=-2$, and now your $P$ is $(1,-2)$. The "$r$" will now be $\sqrt{5}$, and now you can compute ratios like $y/r$.
If you had chosen some other positive $x$, like $x=3$, you would get $y=-6$, $r=\sqrt{45}=3\sqrt{5}$, but the ratios would be unchanged.
Probably in your course $\cos\theta$ is defined to be $x/r$, where $r$ is the distance from your point, in this case $(x,y)=(5,-12)$, to the origin. I computed that distance using the "distance formula," which is basically the Pythagorean Theorem, and got $13$. Then I identified $x$, which is $5$ in this case, and got $\cos\theta=5/13$. Similarly, $\sin\theta$ is defined to be $y/r$. Identify $y$, it is $-12$. Divide by $13$. –  André Nicolas Jun 1 '11 at 19:15