this is section 5.6 graphs of other trigonometric functions and we're going to begin with graphing the secant and the cosecant functions let's remember our definition of a cosecant y equal to the cosecant of X is equal to the reciprocal of the sine function 1 divided by the sine of x so where the sine of x is equal to zero in other words the X intercepts of the sine of x become the vertical asymptotes of the co-seek in the back so let's take a look at this table here where the sine of x is equal to zero we now have the cosecant of X undefined that's because this is 1 divided by the sine of x so these are all of the vertical asymptotes on the cosecant of X this gray graph here this is y equal to the sine of x so if you note that y equal to the sine of x has a 0 at negative pi 0 pi and so forth 2 pi what do you notice that's where the cosecant of X has vertical asymptotes so um the Max and the Min from the sine function correspond to the Min and the max for the cosecant function okay so those are interchanged what used to be a Max on the sine of x is now a minimum on the cosecant of x okay so it's uh very nice let's take a look at y equal to the secant of X again we have that the secant of X is 1 divided by the cosine of x so if this gray graph is y equal to the cosine of x all of the zeros or the X intercepts of the cosine of x now become the vertical asymptotes of the secant of x okay and likewise as you can see where the cosine is equal to zero the secant is undefined so this is where we will have our vertical asymptotes okay the graphs let's just look at a quick summary uh the there is a period of 2 pi for both of these there is no amplitude as you can see these graphs grow without bound um we have the origin as our sine function is an odd function so the cosecant is an odd function and the cosine is an even function so the secant is also an even function okay so we're symmetric with respect to the origin and were symmetric with respect to the y-axis okay so let's go ahead and try to graph this first one y equals negative 2 cosecant of 3x right away you can see that we are going to reflect uh the cosecant of 3x across the x-axis so that's going to be important to start with um let's go ahead and take a look at the amplitude that's the absolute value of negative 2 so that's equal to 2 the period again will take 2 pi and divide it by B which is three so we will have one complete cycle between 0 and 2 pi over 3. okay let's talk about where y equal to 3 sine X has x-intercepts let's do that first note y equal to the C oh it's not the secant sine of 3x has x-intercepts at even multiples of pi over 3. so what does that mean we have for the cosecant graph we have therefore y equal to the cosecant of 3x has vertical asymptotes let me write that at those same places at even multiples of pi over 3. let's go ahead and take a look at those so as you can see I'll use maybe a highlighter this time take a look at that pi over 3 negative pi over 3. and as you can see we have zeroes on the graph that are now vertical asymptotes for the cosecant so very uh nice I believe you can see the amplitude of 2 here and you can see one complete cycle between zero and two pi over three there's one complete cycle okay so now let's go ahead and we could just go ahead and write down how we would use this period of 2 pi over 3. how would we use that we can just do a couple of these so the first thing we would do is we would take one fourth of 2 pi over 3 and we would get pi over 6 and that's not our number one so now what we'll do we do with that we would first start at zero and add pi over six and we would get a point of pi over six negative two okay so we reflected across the x-axis there's the reflection and there's the amplitude Okay and there it is right there that point okay we can do the second one we don't have to do all of them we'll have zero Plus two pi over sixes and that's equal to pi over 3. and that generates a point as you can see that is at a vertical asymptote okay then we'll go ahead and go to three zero plus three pi over six and that will be put vertical asymptote there pi over 2 and that generates a point pi over two two and as you can see that's right here and that would continue on okay so let's take a look at the next graph what's the difference between this graph and this graph you guessed it right here we are going to add one we have a vertical shift up one so we will add one to the Y values of the graph in part a so let's do a few of those points pi over 6 negative two now becomes pi over 2 negative one as you can see right here pi over 2 negative one then if we have that would be uh pi over 6 negative one and you can see it right here and then let's take our next point pi over 6 2 we would have now that's pi over 2 2 my goodness we will have pi over 2 3 and as you can see we're at pi over 2 and we're right here at 3. okay I'll put that point there and I'll put that point there goodness okay so now let's talk about the tangent and the cotangent functions we'll we'll use that same concept of where we have um an x-intercept uh for the sine of x is where we are going to now have our um multiples of Pi so as you can see right here this is y equal to the tangent of X okay so where the sine of x is equal to zero that puts 0 in the numerator and that is why when the sine of x is zero we have an x-intercept on the tangent of graph graph okay now how about a vertical asymptote we have a vertical asymptote where the cosine of x is equal to zero because it's in the denominator and that is um where the cosine is zero so this is a very nice nicely graphed function as you can see we do have an odd function because it's symmetric with respect to the origin each quadrant one point has a mirror image in quadrant three this is actually a feature of being an odd function and the period of the tangent function is pi so one complete period occurs between two consecutive vertical asymptotes so what we're saying is we have right there from pi over 2 to 3 pi over 2 we have one period we have one complete uh cycle so now let's look at the reciprocal of the tangent of X and that's the cotangent of X and these same Concepts apply where the cosine of x is equal to zero that is where we will have our x-intercepts where the sine of x is equal to zero we will have vertical asymptotes okay and you can see that graph here we still have an odd function symmetric with respect to the origin Mirror Image in quadrant three and the period is still pi each one of these is a distance of Pi and so forth okay so now let's look at this summary as you can see we have a period of Pi and we have odd function and you can see the vertical asymptotes multiples of pi over 2 and then also multiples of Pi and these are odd multiples okay when graphing again we still have our period is pi over B the vertical shift is D phase shift C over B and then we will have two consecutive vertical asymptotes and uh between those and we will be solving these inequalities for each one of these graphs one for the tan and one for the coast hand okay so we will always have an x-intercept halfway between the asymptotes okay so let's go ahead and try to graph y equal to the cosine of 4X remember we will have our period and that's pi divided by B and in this case that's 4. okay so we have one cycle one complete cycle I want to say in each period of pi over 4. so now and you can see that right here and so now we could look at this midpoint and let's talk about the x-intercepts at these midpoints so we'll take 0 plus 5 pi over four and divide by 2 and that comes out to be pi over 8 and that is an x-intercept right here and you could do that on each interval okay let's take a look at this next graph y equals 4 times the tangent of 2x plus pi over 3 when we have these additions in there it can be a little bit tricky so we're going to go ahead and write this out we have negative pi over 2 and then we have 2X plus pi over 3 in between negative pi over 2 and pi over 2. okay so now let's subtract pi over 3 from every part of this three-part inequality and we come up with negative 5 pi over six and pi over six and then dividing each part by two and so we have one complete cycle between negative pi 5 pi over 6 and pi over 12. and you can see that right here between pi over 12 and negative 5 and that should be 12. right here okay and another one there so all of those are vertical asymptotes and you can see where we have uh one complete cycle we could also find those x-intercepts let's try that our x-intercepts are at our midpoint of those vertical asymptotes so negative 5 pi over 12 plus pi over 12 all divided by 2 and we will get negative 4 pi over 12 all divided by 2. and if you do that we will have we we can multiply by a half and so that is not written very good is it I'll try to write a little neater and maybe I could bring this over here and then just right across so now this is equal to negative 4 pi over 12 all divided by 2 negative 4 pi over 12 divided by 2 is multiplied by 1 dividing by 2 is the same as multiplying by one half so we would actually have and then that 12 and that 2 can cancel and we would have a negative pi over 6 and these are our X intercepts Okay negative pi over 6 and we'd get pi over 3 we could do that for each uh interval between the vertical asymptotes okay and now in number four it is not the same graph but we are we still have a tangent graph and the very first thing you should notice is we have a reflection across the x-axis okay so what do we normally have for the tangent of X is now going to be looking like this this is the opposite of the tangent of X maybe I better put y equals so you know that that Dash that that negative sign is there and it's not a dash so this is the reflection and as you can see we will go up to a vertical shift going up two and uh let's go ahead and put that interval in between negative pi over 2 and pi over 2. so now multiply everything by four and then dividing by three I think I forgot an X in here and then we're dividing uh everything by 3 pi and we'll have negative two-thirds is less than x less than two-thirds and so as you can see right here we have negative two-thirds and two-thirds okay so we have one complete cycle there and let's go ahead and see if we can come up with our x-intercept and we could do that before we move up two units so before moving up before shifting up two here's how we will find our x-intercepts and we'll just find one of them so we'll take the midpoint of negative two-thirds and add that to two-thirds and divide that by two but that's equal to zero so that is why we have this zero zero so here we will move up to two units for uh the graph of this function with the plus two okay and that completes this section