# Need help drawing this graph-AL Maths

https://www.quora.com/profile/Bravewarrior/How-is-this-the-graph-of-y-cosec-2x-60-I-have-also-attached-a-pic-of-cosec-x

So the first pic shows the graph of y=cosec(2x+60). I also attached a pic of y=cosec(x). If the coordinates on cosec(x) are (90,1),(180,0),(270,-1) and (360,0) then how does the graph end up looking like the one in the mark scheme which I attached? Any help would be great!
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/How-is-this-the-graph-of-y-cosec-2x-60-I-have-also-attached-a-pic-of-cosec-x

So the first pic shows the graph of y=cosec(2x+60). I also attached a pic of y=cosec(x). If the coordinates on cosec(x) are (90,1),(180,0),(270,-1) and (360,0) then how does the graph end up looking like the one in the mark scheme which I attached? Any help would be great!

cosec(x) can never be 0 as its 1/sin so its >=1 or <= -1. Can you sketch sin or cosec of
sin(x)
sin(2x)
sin(x+60)
sin(2x+60)
then it should be clear? The first local min on the graph should be (15,1) and the first local max should be (105,-1) so the labels are a bit off.
(edited 2 months ago)
Original post by mqb2766
cosec(x) can never be 0 as its 1/sin so its >=1 or <= -1. Can you sketch sin or cosec of
sin(x)
sin(2x)
sin(x+60)
sin(2x+60)
then it should be clear? The first local min on the graph should be (15,1) and the first local max should be (105,-1) so the labels are a bit off.

I think I kind of get it? But thank you for helping!
Why not sketch what you think it should be and put the labels on and upload?
If you kind of get it, what are you unsure about?
Original post by mqb2766
Why not sketch what you think it should be and put the labels on and upload?
If you kind of get it, what are you unsure about?

I am stuck on this basically:

If it is y=cosec(2x+60) then would I not divide all x-coordinates by 2 and also subtract 60 from each one? The y-values would remain the same? So (90,1) would become (-15,1), (180,0) would be (30,0), (270,-1) would be (75,-1) and (360,0) would be (120,0)? As the range is larger then the graph would repeat and so on and so forth...

Original post by pigeonwarrior
I am stuck on this basically:

If it is y=cosec(2x+60) then would I not divide all x-coordinates by 2 and also subtract 60 from each one? The y-values would remain the same? So (90,1) would become (-15,1), (180,0) would be (30,0), (270,-1) would be (75,-1) and (360,0) would be (120,0)? As the range is larger then the graph would repeat and so on and so forth...

Firstly you seem confused by cosec and sin and secondly the basic transformations. For both of these, if in doubt stick some (critical) numbers in.

So by an inspired choice of x values (which you should think about /understand)
x=0 so sin(2x+60)=sin(60)=sqrt(3)/2 and cosec(60)=1/sin(60)=2/sqrt(3)
x=15 so sin(2x+60)=sin(90)=1 and cosec(90)=1/sin(90)=1
x=60, so sin(2x+60)=sin(180)=0 so cosec(180)=1/sin(180)~inf
.... Try it at x=105, 150, ...

To understand the transformation, you have
cosec(z) = 1/sin(z)
where z=2x+60=2(x+30) or
x = z/2 - 30
so the origin z=0 is translated to x=-30 (so vertical asyptote etc) and the period of 360 for z becomes 180 for x so vertical asymptotes for cosec every 90 and similarly for the local min and max.

Again, try sketching what you think and upload.
(edited 2 months ago)
Original post by mqb2766
Firstly you seem confused by cosec and sin and secondly the basic transformations. For both of these, if in doubt stick some (critical) numbers in.

So by an inspired choice of x values (which you should think about /understand
x=0 so sin(2x+60)=sin(60)=sqrt(3)/2 and cosec(60)=1/sin(60)=2/sqrt(3)
x=15 so sin(2x+60)=sin(90)=1 and cosec(90)=1/sin(90)=1
x=60, so sin(2x+60)=sin(180)=0 so cosec(180)=1/sin(180)~inf
.... Try it at x=105, 150, ...

To understand the transformation, you have
cosec(z) = 1/sin(z)
where z=2x+60=2(x+30) or
x = z/2 - 30
so the origin z=0 is translated to x=-30 so vertical asyptote etc) and the period of 360 for z becomes 180 for x so vertical asymptotes for cosec every 90 and similarly for the local min and max.

Again, try sketching what you think and upload.

Hello, I managed to upload a pic of what I've sketched under the two images I had previously posted here Need help with this question - Quora .
I didn't sketch the whole graph for that interval but this is what I have so far...
It is a bit different to the one in the mark scheme and also I noticed that in the mark scheme they called it y=cossec(2x+90) but the question asked me to draw y=cosec(2x+90) so not sure if that has something to do with it?
Original post by pigeonwarrior
Hello, I managed to upload a pic of what I've sketched under the two images I had previously posted here Need help with this question - Quora .
I didn't sketch the whole graph for that interval but this is what I have so far...
It is a bit different to the one in the mark scheme and also I noticed that in the mark scheme they called it y=cossec(2x+90) but the question asked me to draw y=cosec(2x+90) so not sure if that has something to do with it?

So is it cosec(2x+90) or cosec(2x+60) you want to sketch? Your sketch is for the latter and you have the period correct (180, so asymptotes and min/max every 90 which is due to the 2x) but the location isnt correct. The min/max/asymptotes are explained in the previous post using
x = z/2-30
so asymptotes occur at z=0,180,360,... or x=-30,60,150... and similarly for min/max at x=15,105,195,...

Not sure exactly how you come up with your numbers.
(edited 2 months ago)
Original post by mqb2766
So is it cosec(2x+90) or cosec(2x+60) you want to sketch? Your sketch is for the latter and you have the period correct (180, so asymptotes and min/max every 90 which is due to the 2x) but the location isnt correct. The min/max/asymptotes are explained in the previous post using
x = z/2-30
so asymptotes occur at z=0,180,360,... or x=-30,60,150... and similarly for min/max at x=15,105,195,...

Not sure exactly how you come up with your numbers.

Yup it is for cosec(2x+60) and I get now where I've gone wrong! Thanks for all your help! π