**3. How do the graphs of sine and cosine relate to each of the others?**

a) Tangent?

First, let's sketch the sine and cosine graphs.We know that sine and cosine both have a period of 2π with sine starting and ending at 0 while cosine starts and ends at its amplitude. Looking back to the Unit Circle, we know that sine is positive in the 1st quadrant (0 to π/2) and in the 2nd quadrant (π/2 to π) and negative in the 3rd and 4th quadrants (π to 2π). Cosine is positive in the 1st and 4th quadrants, but negative in the 2nd and 3rd quadrants.

Now according to the ratio identities, tangent is equal to sine/cosine. So we would be correct to assume that if sine or cosine was negative, then tangent would also be negative. Following this train of thought, that would mean that if both sine and cosine were positive or negative, then tangent would be positive. Therefore, since sine and cosine are both positive in the 1st quadrant (0 to π/2), then tangent would also be positive. Or since cosine is negative while sine is positive in the 2nd quadrant (π/2 to π), then cosine would cause tangent to be negative as well. In the 3rd quadrant, both sine and cosine is negative, so that cancels out to make tangent positive. In the 4th quadrant, cosine is positive, but sine is negative, causing tangent to be negative.

In addition, since tan= sin/cos, then the tangent graphs would have asymptotes whenever cosine=0, making the ratio undefined. According to the Unit Circle, we know that cos=0 when the x value is 0, so at π/2 (90*) and 3π/2 (270*). The asymptotes then keep repeating itself at the same intervals since the domain goes on and on from left to right. Tangent will never be able to touch and cross the asymptotes, no matter how close it gets to it.

b) Cotangent?

Going back to the sine and cosine graphs, cotangent is going to be very similar to the tangent graph. Cotangent, since it's the reciprocal of tangent, is equal to cosine/sine.So we would be correct to assume again that if sine or cosine was negative,
then cotangent would also be negative. As a result, that would mean that if both sine and cosine were positive or negative,
then cotangent would also be positive. Therefore, since sine and cosine are
both positive in the 1st quadrant (0 to π/2), then cotangent would also be
positive. Or since cosine is negative while sine is positive in the 2nd
quadrant (π/2 to π), then cosine would cause cotangent to be negative as
well. In the 3rd quadrant, both sine and cosine is negative, so that
cancels out to make cotangent positive. In the 4th quadrant, cosine is
positive, but sine is negative, causing cotangent to be negative.

However, since cot= cos/sin then the cotangent graphs would have asymptotes whenever sine=0, making
the ratio undefined. According to the Unit Circle, we know that sin=0
when the y value is 0, so at 0 or 2π (360*), and π (180*). The asymptotes
then keep repeating itself at the same intervals since the domain goes
on and on from left to right. Cotangent will never be able to touch and
cross the asymptotes, no matter how close it gets to it. Furthermore, because of the different asymptotes that act as boundary, cotangent graphs go "downhill" instead of "uphill" like tangent graphs do.

c) Secant?

How do secant and cosine relate? Well, looking at the cosine graph alone, we know that cosine is positive in the 1st and 4th quadrants, but negative in the 2nd and 3rd quadrants. Since secant is the reciprocal of cosine and correlates to it, secant is also positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants.

Also, since secant is the reciprocal of cosine, then the secant graph is basically the cosine graph flipped. For example, since the the slope of the cosine graph is negative in the 1st quadrant, making the values very small fractions. So if we were to take the reciprocal of that for secant, we would end up with these huge numbers.

However, since sec=1/cos, then there would be an asymptote whenever cos=0, making the ratio undefined. We know that cos=0 when the x value is 0, so at π/2 (90*) and 3π/2 (270*) there would be an asymptote for secant, which repeats themselves again forever at the same intervals. The secant graph then is at the peak of each cosine mountains and at the bottom of each cosine valley.

d) Cosecant?

How does cosecant relate to sine? Looking at the sine graph alone, we know that sine is positive in the 1st and 2nd quadrant, but negative in the 3rd and 4th quadrant. Since cosecant is the reciprocal of sine (csc= 1/sin), then cosecant would be positive wherever sine is positive and negative wherever sine is negative. Cosecant would also have to touch the peaks of the sine mountains and the bottom of the sine valleys. Then, it spreads out from both sides.

Then, we know that that csc= 1/sin. So cosecant would basically the sine graph flipped. However, cosecant will have asymptotes wherever sin=0, since that would make the ratio undefined. Sin=0 at 0 or 2π (360*), and π (180*). The cosecant graphs would spread out from the peak/valley and approach as close it can possible can to the asymptotes (because they keep on repeating over and over again at the same intervals) on either side but will never touch or cross them, instead the y-value would keep increasing/decreasing (going to +/- infinity).

Also, since secant is the reciprocal of cosine, then the secant graph is basically the cosine graph flipped. For example, since the the slope of the cosine graph is negative in the 1st quadrant, making the values very small fractions. So if we were to take the reciprocal of that for secant, we would end up with these huge numbers.

However, since sec=1/cos, then there would be an asymptote whenever cos=0, making the ratio undefined. We know that cos=0 when the x value is 0, so at π/2 (90*) and 3π/2 (270*) there would be an asymptote for secant, which repeats themselves again forever at the same intervals. The secant graph then is at the peak of each cosine mountains and at the bottom of each cosine valley.

d) Cosecant?

How does cosecant relate to sine? Looking at the sine graph alone, we know that sine is positive in the 1st and 2nd quadrant, but negative in the 3rd and 4th quadrant. Since cosecant is the reciprocal of sine (csc= 1/sin), then cosecant would be positive wherever sine is positive and negative wherever sine is negative. Cosecant would also have to touch the peaks of the sine mountains and the bottom of the sine valleys. Then, it spreads out from both sides.

Then, we know that that csc= 1/sin. So cosecant would basically the sine graph flipped. However, cosecant will have asymptotes wherever sin=0, since that would make the ratio undefined. Sin=0 at 0 or 2π (360*), and π (180*). The cosecant graphs would spread out from the peak/valley and approach as close it can possible can to the asymptotes (because they keep on repeating over and over again at the same intervals) on either side but will never touch or cross them, instead the y-value would keep increasing/decreasing (going to +/- infinity).

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