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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
up vote 2 down vote accepted

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: On your second question, yes, correct, right reasoning.

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.

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I don't understand what happened between 5,-12 and the r= part – Adam Jun 1 '11 at 19:08
r is the distance from the point to the origin. – The Chaz 2.0 Jun 1 '11 at 19:11
Alright I got it it now, x squared plus y squared is equal to radius squared. I forgot completely. Anyways anyone have any tips on how to utilize this archaic tome of masochism that is my math book? It has no answers in it so I spend hours doing something wrong and never realize it. – Adam Jun 1 '11 at 19:14
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
@Adam, if you don't like your math book, get another one. The books in the Schaum's Outline series have hundreds of completely worked-out examples. – Gerry Myerson Jun 2 '11 at 3:40

Question 1:

**A quick note before we start, remember that the unit circle is 360 degrees. When something asks for the "least possible positive measure", it's just asking for something between 0 and 360 degrees.

Let's break it down into parts.

I find it's best to visualize what we're actually doing. Therefore, we're first going to draw the right triangle the question gives you.

  1. Plot the point (5, -12) on a graph.
  2. Draw a line from the origin, also known as point (0, 0) on your graph, to the point you just plotted at (5, -12). This line you just drew gives you the hypotenuse side of the triangle you're about to draw.
  3. Now draw a line from the origin (0, 0) to point (5, 0)
  4. Now draw another line from point (5, 0) to the original point you plotted (5, -12).

At the moment you should have a nice right triangle drawn. You may have noticed that you also have the side lengths of your triangle, given by the x and y values. The x-value is 5 and the y value is -12. With that we can now find the length of the hypotenuse.

Using the Pythagorean Theorem we know that: a-squared + b-squared = c-squared. So now we just plug in the numbers.

5(5) + -12(-12) = 25 + 144 = 169


c-squared = 169

We now take the square root of 169 to come up with 13. We now have the length of the hypotenuse at 13. Now we have all of the values we need to determine our trigonometric functions. I don't know if you're familiar with this:

Soh Cah Toa Cho Sha Cao

It basically means that:

Sine = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tangent = Opposite / Adjacent
Cosecant = Hypotenuse / Adjacent
Secant = Hypostenuse / Adjacent
Cotangent = Adjacent / Opposite

Recall that the opposite value is the same thing as the y-value, which in this problem is -12, and the adjacent value is the x-value, which in this problem is 5, and we determined the hypotenuse value to be 13 or:

Opposite = -12
Adjacent = 5
Hypotenuse = 13

Now that we've defined the six trigonometric functions, and have our values we just plug in the numbers:

Sine = -12/13
Cosine = 5/13
Tangent = -12/5
Cosecant = 13/-12 or -13/12
Secant = 13/5
Cotangent = 5/-12 or -5/12

Question 2:

Remember that a ratio is just a fraction. So when it's asking if the ratio is positive or negative, that's just another way of asking "is the fraction positive or negative?".

So if point (x,y) is in quadrant 2, we know that the x-value will be positive and the x value will be negative.

Going back to our Soh Cah Toa Cho Sha Cao we can just plug in values. Remember that the y value is the Opposite and the x-value is the Adjacent.

Sine = y/h making this a positive ratio
Cosine = -x/h making this a negative ratio
Tangent = y/-x or -y/x making this a negative ratio
Cosecant = h/y making this a positive ratio
Secant = h/-x or -h/x making this a negative ratio
Cotangent = -x/y making this a negative ratio

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