Unit T Big Question # 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill? 

• Looking at the Unit Circle and ASTC, we see that tangent is positive in the first quadrant, negative in the second, positive in the third, and negative in the fourth. Now if we take the unit circle and make it so that it is on a graph it would be like this: 

Tangent is sine over cosine and we know that there is an asymptote whenever the value is undefined. Since tangent is sin/cos then the value will be undefined whenever cosine equals zero and that happens at pi/2 and 3 pi/2. 

                                                 pi/2 ( 0, 1) -------- y/x= 1/0=undefined

                                                 3 pi/2 (0,-1)-------y/x= -1/0= undefined

So we have an asymptote at pi/ 2 and 3 pi/2 which is represented by the picture below:

Now that we have the asymptotes we have the barrier to each graph. So we had said that tangent is positive in the first quadrant so that means that between 0 and pi/2 the tangent curve will be going up above the x-axis towards the asymptotes but not touching it. Between pi/2 and pi tangent is negative so the curve will be below the x-axis going up towards the asymptote. If we do this for all the quadrants we get a result of the following graph, which shoes the final tangent graph:

* This is why the tangent graph goes up and it is really based on the placement of the asymptotes that guide how we draw tangent graphs. 

• For cotangent the same process is taken to explain why cotangent graphs are downhill.

 Looking at the unit circle cotangent is positive in the first quadrant, negative in the second, positive in the third, and negative in the fourth. So again like tangent we draw the unit circle on a plane. This type however cotangent is cosine over sine. So this time the asymptotes will lie whenever x/y is undefined.This only happens at pi and 0 or 2 pi.

                                               pi ( -1,0)------x/y= -1/0 = undefined

                                              0 / 2 pi (1,0)------x/y= 1/0= undefined 

*So we have asymptotes at pi and 0 and 2 pi. 

Now we set our barriers we can draw the graph. Between 0 and pi/2 cotangent is positive so it will be above the x-axis only this time the curve will be going down because the it is framed by the asymptote. We do this for all the quadrants and in result get the following graph,which shows the final cotangent graph:

Citations for images:

http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphtrigfunctions.html

http://demo.activemath.org/ActiveMath2/search/show.cmd?id=mbase://LeAM_calculus/curves/ex_cot_pole