 So let's take a look at trigonometric functions beyond the right angle. So let's consider this trigonometry in the unit circle. Suppose our angle has been moved into standard position. So remember that's going to have the vertex at the origin, one side on the positive x-axis, and other side wherever it ends up. Now let's consider a unit circle centered at the origin. That's a circle with radius 1 centered at the origin. The terminal side of the angle will intersect the unit circle at some point b. And since we're on a set of coordinate axes, we can say that this point b has coordinates x, y. Now let's consider this coordinate a little bit more carefully. If the point b has coordinates x, y, then this vertical line, b, c, has length y. Likewise, this horizontal line, a, c, has length x. And a useful rule in mathematics is look for the right triangle. Since b, c is vertical, that means that this angle b, c, a is a right angle, and so a, b, c is a right triangle. Now we know a, c is x and b, c is y, but what about this third side, a, b? Well remember, that's the distance between the center of the circle and the point on the circle, and since this is a unit circle, the radius a, b has length 1. Now since a, b, c is a right triangle, we can talk about the sine, cosine, and tangent of this angle a. So the sine of a is going to be the length of the opposite side divided by the length of the hypotenuse. That opposite side has length y, and the hypotenuse, because it's the radius of the unit circle, has length 1. So the sine is y over 1, or just y. Similarly, the cosine of the angle is the adjacent over the hypotenuse, and that's going to be x over 1, or just x. And finally, the tangent is going to be the opposite over the adjacent, and that's going to be y over x. So now let's change our viewpoint slightly. Suppose angle a is in standard position. Let b be the point where the terminal side intersects the unit circle with coordinates x, y. Then the sine of a is equal to y, the cosine of a is equal to x, and the tangent of a is equal to y over x. And what's useful to keep in mind here is while this was determined by assuming that this angle a was an acute angle, we can still find these values x, y, and y over x, whatever the measure of the angle a is. And this allows us to introduce the following. Let theta be an angle in standard position and let its terminal side intersect the unit circle centered at the origin at a point with coordinates x, y. Then the sine of theta is y, the cosine of theta is x, and the tangent of theta is y over x. Now we have to find sine, cosine, and tangent for any angle theta. But the other trigonometric functions can also be defined in terms of sine, cosine, and tangent. And so this allows us to define the rest for any angle theta. secant theta is whatever cosine. cosecant is whatever sine. And cotangent is whatever tangent. So let's go back to our nonsense question, what is the sine, cosine, and tangent of 120 degrees? And so what we might do is we might start by drawing an angle with a measure of 120 degrees in standard position. This angle meets the unit circle at some point x, y. Since we don't know the value of x or y, we should. We should say what we can about it. Remember, it's better to speak the truth that you know than to not say anything at all. So remember that if we put our angle A in standard position and B is the point where the terminal side intersects the unit circle with the coordinates x, y, then the sine of the angle is y, the cosine is x, and the tangent is y over x. Now if we look at where this point is, we see that because we're over here on the left side of the y-axis, x is less than zero. But equals means replaceable. If cosine of 120 is equal to x, then any time we see x, we can replace it with cosine of 120. And so that tells us that the cosine of 120 degrees is less than zero, it's negative. Likewise, we see that this point is above the x-axis, which tells us that y is greater than zero. And since sine of 120 degrees is equal to y, equals means replaceable, any time you see y, you can replace it with sine of 120. And so we know that sine of 120 degrees is greater than zero, it's positive. And finally, let's consider this y over x because y is positive and x is negative, y over x is going to be less than zero, it's going to be negative. And equals means replaceable, y over x is the same as tangent 120, and so we know tangent will be less than zero, it will be negative. While we could stop here, there's a useful thing to remember. Knowledge is power. Answer the related questions. In this case, we've learned something about the cosine, and tangent of 120 degrees. But we know that the sine, cosine, and tangent are related to the secant, cosecant, and cotangent. They are in fact the reciprocals of these functions. And so once we know the cosine of 120 is negative, then we know that the secant, which is one over cosine, is going to be negative. Likewise, we know that the sine is positive, and so the cosecant is going to be positive. The tangent is negative, and so the cotangent is negative. What if we go the other way? Suppose we know that the sine of an angle is positive and the tangent is negative. What can we say about the angle? Well, again, if we put the angle in standard position, then its terminal side intersects some point B on the unit circle with the coordinates x, y, where the sine is y, the cosine is x, and the tangent is y over x. Since the sine is positive, then we know that the sine is greater than zero, and equals means replaceable. y and sine theta are the same thing, so we can replace sine theta with y, and we find that y is greater than zero. Now we also know that tangent is negative, and since tangent equals y over x, we can replace tangent with y over x, and we know that y over x is negative. Now we know y is greater than zero, y is positive, so in order for y over x to be negative, x must be less than zero. That's because in order for a quotient y over x to be negative, y and x have to have opposite SIGN signs. So if y is positive, x has to be negative, and equals means replaceable, so incidentally this tells us that cosine also has to be negative. As far as the angle is concerned, we're at a point where y is positive, and x is negative, and that puts us someplace in the second quadrant. And that means that our angle has to be somewhere between a quarter of a turn and a half turn, someplace between 90 degrees and 180 degrees.