I understand that the definition of a tangent of an angle is (opposite/adjacent)...So if i take the tangent of a right angle, shouldn't i get 2 answers?:bugeye:

My teacher and my calculator say its undefined, and my teacher brought the unit circle into it, but I dont understand it. Help please? Thanks...

draw a right triangle. make one of the other angles also a right angle. what are you left with? the opposite side of this "triangle" still has some height. but the adjacent side of this triangle has disappeared! so the tangent of a right angle is some number (height) divided by zero! hence undefined.

If you are familiar with the graph of a tangent function, you should notice that at pi/2 the value of the function is infinity. The general notion of the tan-function is that as the angle increases from 0 to pi/2, tan increases without bound. So if you grab any right-angled triangle (R-A-T) and stick say a pi/4 (45deg) angle in it, if you increase that angle to 89.999deg what you have in fact is a very thin triangle where two of the sides are almost 90 degrees each. Of course if that happened (two right angles with three sides) one of the sides disappears and the triangle becomes a straight line.

Now as you increase the angle from 45deg to 89.999deg what is happening to our tan-function? Well, the definition of a tan-function is the length of the opposite side divided by the adjacent side. If you tried to use the base and the height in our 89.999deg triangle you would have tan(q) = b/h. As we worked out before the height (h) of the triangle must be very small (in fact, what is the height of a 1-dimensional line? Zero!) So if we had two 90deg angles in a triangle we have a line with zero height.

Thus, tan(q) = b/0 = undefined (you cannot divide a number by 0 at all, it just doesn't make sense!)

Alternatively, you could consider that the opposite side goes to infinity as the angle approaches 90 degrees.

Ok... suppose we just have a nice 30-60-90 triangle. In a sipmle case like that, it still can be done though cant it?

I understand what youre talking about about t virtual 90 degree angles, but what about the other case i brought up?

thanks:D

I understand what you are having difficulty with now! :)

Well firstly we need to define exactly what 'taking the tangent of an angle' really is!

Obviously, tangents are a little bit more involved than sine and cosine. Here I could quickly say that tan is defined as the ratio of the opposite side over the adjacent side, so if you take the right angle there is no hypotenuse anymore becuase it HAS to be either the adjacent or tangent you see. If you extend that angle to more than 90 degrees you cannot use Pythagoras/Trig anymore and you have to use the sine or cosine rules.

However, if you don't believe that keep reading...

Tangent can be restated as sine/cosine. So we are really trying to define what 'taking the sin/cos of an angle' is. What I am saying is that whenever tan pops up in trig, you have to think of it as sin/cos. Tan was only introduced to trig as another way to write sin/cos, that's it!

Now here I could quickly say that tan(pi/2) = sin(pi/2)/cos(pi/2) = 1/0 = Error!!! Undefined! Cannot devide by zero. We say tan(pi/2) is undefined at pi/2. Simply because sin(pi/2)/cos(pi/2) doesn't work.

However, if you don't believe that keep reading...

If you really want to have a good understanding of this concept you will have to know a bit about functions (a lot actually!). But I will keep it simple.

Imagine (or draw) the graph of the sine function on a set of axes with intervals of (pi/2) drawn on the x-axis and intervals of 1 drawn on the y-axis. The curve will oscillate between 1 and -1 starting from 0 at 0 and reaching the first maximum at (pi/2). It will continue to oscillate in this manner but we are only concerned with what happend near (pi/2) right!

Now plot the cosine function and notice that it approaches 0 as x approaches (pi/2).

You should realise that tan (the function) is simply a division of the sine functions and cosine functions.

Now at (pi/2) f(sin) = 1 and f(cos) = 0. We can by the algebra of functions divide both these functions algebraically. So that f(sin)/f(cos) = 1/0. Hence and error occurs.

Now try to plot this. Take a point somewhere very close to (pi/2) but not at (pi/2) say 89.999degrees. sin(89.999) = 0.99999999999 and cos(89.999) = 0.000017453. Thus tan(89.999) = 0.999999999/0.000017453 = 57296.7... This is a very big number, and imagine that we keep adding 9's to sin and cos. The number approaches infinity doesn't it?! Another way we tend to write this is as cos(x) and sin(x) approaches (pi/2) the denominator approaches 0 and the numerator approaches 1.

sin(x)/cos(x) as x goes to (pi/2) = 1/0 (or in someways infinity)

Now of course we cannot have function existing at infinity let alone defined there, the best we can do is define the function near infinity but not at infinity. (Incidentally that is where the domain of sine is (-inf, +inf) notice the 'round' parentheses, meaning "not included in the interval". This is very important when talking about functions as this is very different from a 'square' bracket which means the endpoints are included!)

So since tan is simply a combination of two other functions which do not exist at pi/2 we say tan has a 'vertical asymptote' at pi/2.

A tangent at 90degrees (pi/2) is hence undefined!!

I hope this helps. Cheers. Ben.