graphical representation of trig functions

Question

I'm currently learning the unit circle definition of trigonometry. I have seen a graphical representation of all the trig functions at khan academy.

I understand how to calculate all the trig functions and what they represent. Graphically, I only understand why sin and cos is drawn the way it is. I'm having trouble understand why tangent, cotangent, secant, cosecant are drawn the way they are.

Can someone please provide me with some intuitions.

Answer 1

While this figure is elegant in its way, it has a little bit too much going on for the beginning student.

Lets start with a simpler figure.

Our two triangles are similar. The smaller triangle has side lengths $(\cos x, sin x, 1)$

Multiply all three by the same ratio $\frac{1}{\cos x}$ we will get the side lengths of the similar triangle.

$(1,\tan x, \sec x)$

All of the "co" functions are "flipped" across the line $x=y$

Relating these figures to the first one, I will point out that these are congruent triangles.

Answer 2

It would be very helpful if you connected the center of the unit circle to the point of tangency.

Notice that this segment has length one and is perpendicular to the tangent line.

Now there are many similar right triangles to be found and the proprtionality relations gives you the trig functions as indicated in the graph.

It is very time consuming but the result is interesting.

Answer 3

At least in the first quadrant of the unit circle, the tangent of the angle is equal to the length of the tangent segment connecting the point on the circle to the x-axis.

So in your image, imagine a radius being drawn from the origin to the green point on the circle's circumference (let's call it $P$). Call the angle between this radius and the positive x-axis $\theta$.

In terms of the x and y coordinates of $P$, $\tan \theta = \frac y x$.

If you think more in terms of right triangle trig and "SOH CAH TOA", then $\tan \theta = \frac{\text{ opposite}}{\text{ adjacent}} = \frac{\text{ length of tangent segment }}{\text{ length of radius }} = \frac{\text{ length of tangent segment }}{1} = \text{ length of tangent segment }$.

Answer 4

If you complete the diagram with all the right triangles, you would notice all of them are actually similar, meaning pairs of corresponding sides produce equivalent ratios. I’m not exactly sure how to explain this considering there aren’t any points in your diagram, so here’s a link to the proof I used.
Trigonometric Ratios on a Unit Circle