BQ#3 – Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each of the others?  Emphasize asymptotes in your response.
Sine and cosine are found in the ratio identities of the rest of the four trig functions; they affect each one of those trig functions because of this.
Tangent
With tangent, we know that tangent is equal to sine over cosine. Now, just by knowing the values of sine and cosine, we can assume what tangent will be on the graph. Look at the visual below to better your understanding.

(https://www.desmos.com/calculator/cwdr1eyszr)

        If you look at the red highlighted area, that is 0 to pi/2, so basically the first quadrant. Even without looking at this graph, we know that sine and cosine are positive, so tangent must be positive as well (positive divided by a positive is a positive). And you can see this translate on the graph. If you look at the green cosine values and the red sine values, you can tell they are above the x-axis, meaning they are positive. And, we determined that since those to trig functions are positive, tangent must be positive, which it is (the orange line). You can see the same thing going on in the next parts of the graph. In the "Quadrant II", on the graph, you can tell that cosine is negative (as the green values fall below the x-axis) and sine is positive (as the red values are above the x-axis). And because of our ratio identity of tan(x)=sin(x)/cos(x), we know that a positive divided by a negative makes a negative. That is why the tangent values fall below the x-axis this time. The 3rd quadrant, again, look at our sine and cosine values: they are both negative. So, a negative divided by a negative is a positive: tangent values are positive. Finally, you should be understanding the relationship that tangent, cosine, and sine have. The last part of the period, cosine is positive (values are above the x-axis) and sine is negative (values below the x-axis); tangent is negative. 

         Now, that you know why the tangent graph looks as it does (through the understanding of its relationship with sine and cosine), we need to deal with another part of these graphs: the asymptotes. The asymptotes are not just there because we say so. There is a reason behind it. Remember when we talked about how tan(x)=sin(x)/cos(x)? We are using this identity again. We know that asymptotes are created when we have an undefined value. We get undefined values when we divide by 0. So, knowing that... we look to our ratio identity and see that cosine must equal 0 in order to get an undefined value (because we are dividing by cosine). How do we get cosine to equal 0? Well, we know that cosine is essentially the x-values of certain points in our unit circle. The only parts of the unit circle that have 0 as an x-value are pi/2 (0,1)and 3pi/2 (0, -1). Therefore, these marks on the graph are where we have asymptotes! Look at the graph above and see where it seems like the tangent graph discontinues; do you notice that pi/2 and 3pi/2 are where we don't really see the tangent graph go? That is because of the asymptotes! 

Cotangent

Cotangent has sine and cosine in its ratio identity as well. Cot(x)=cos(x)/sin(x). 

(https://www.desmos.com/calculator/cwdr1eyszr)

Cotangent has the same thing going on as we saw in tangent. Remember to look at the values of the sine and cosine graphs and relate them to cotangent. For example, see how sine and cosine values are positive, so we have cotangent as positive. However, in the green section, we see that sine (the red graph) has positive values and cosine (green graph) has negative values; we can deduce that cotangent is negative because a negative divided by a positive is a negative. 

Because cotangent has a different ratio than tangent, we are going to be having some different ratios. The ratio identity of cotangent is cot(x)=cos(x)/sin(x). Again, asymptotes are undefined values, meaning they are divided by 0. So, because sine is our denominator in this ratio, sine must be equal to 0 in order to find asymptotes. Where on our unit circle do we have sine=0? Sine is y/r, basically the y values of the unit circle. So, we know the only parts of the unit circle where the y values are 0 are 0 (1,0) and pi (-1,0). Knowing this, we know that our asymptotes are at 0 radians and pi. Look at the graph above and check to make sure. Notice how at 0 radians, a new period begins, as if we lifted our pencil to create a new part of the graph. The same is seen at the pi mark. 

Secant

Secant is a little different from tangent and cotangent. Its ratio identity is 1/cos(x).

(created at https://www.desmos.com/calculator)

We have the same idea though, as we saw in the previous graphs. If cosine is negative, then secant values must be negative as well. If cosine is positive, the secant graph must be positive as well. You can see this in the graph. The cosine values (the black graph) are positive in the red shaded part of the graph; and look at that, the secant graph (the red graph) is positive in that section as well. You can see the trend in the following sections/quadrants as well. It is good to remember that the sine values in here do not have any effect on the secant graph because it is not part of its ratio (*disregard the orange sine graph in the picture above*). 

The asymptotes of secant are the same as we saw in tangent. Tangent had cosine as the denominator as well. So, we know that at pi/2 (0,1) and 3pi/2 (0,-1) there are going to be asymptotes for this secant graph. 

Cosecant

Cosecant has the same idea as secant just reversed, like we saw with cotangent and tangent. Cosecant's ratio identity is csc(x)=1/sin(x). 

(created at https://www.desmos.com/calculator)

Again, we remember that if sine is positive on the graph, then cosecant must be positive as well, because 1 divided by a positive is always a positive--and vice versa. You can see in the red shaded part, the sine graph (the orange line) is positive and the cosecant graph (the red line) is positive as well. It is good to remember that cosine has no effect on cosecant because it is not found in the ratio (*therefore, disregard the cosine graph in the picture above*). 

Cosecant has its asymptotes wherever sine is 0 (because we are dividing by sine in this ratio). Looking back to our cotangent graph, we remember that sine was 0 at 0 radians and pi (because of the coordinates of (1,0) and (-1,0)). So, we will have cosecant's asymptotes at 0 radians and pi. That is why, if you look at the picture above, you see that the graph seems to cut out at 0 radians and pi.