# Graphing trig functionsWatch

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#1
Hi,
I am working on "Graphing Functions" and have come across a term I dont understand. I'm sure its something simple but new to me. Have attached the question (and sample answer) and written in red (under part 5) the bit i dont get.
Feel free to explain what they mean in part 7 also....dont let me stop you..hehehe!!
Would be great to hear from those in the know!

Many thanks
0
1 week ago
#2
(Original post by id0ntkn0w)
Hi,
I am working on "Graphing Functions" and have come across a term I dont understand. I'm sure its something simple but new to me. Have attached the question (and sample answer) and written in red (under part 5) the bit i dont get.
Feel free to explain what they mean in part 7 also....dont let me stop you..hehehe!!
Would be great to hear from those in the know!

Many thanks
Assuming I have the right bit
f'(x) = 1 + cos(x)
which equals zero when cos(x) = -1, or x = (2k+1)pi where k is an integer
1
1 week ago
#3
The 2k+1 bit refers to every odd multiple of pi.
1
#4
(Original post by mqb2766)
Assuming I have the right bit
f'(x) = 1 + cos(x)
which equals zero when cos(x) = -1, or x = (2k+1)pi where k is an integer
Yes, thats it. Im afraid I dont quite understand why it has stationary points at (2k+1)pi?
Also, in point 7, i dont understand after it says "Also for all integers K". Ive included a photo of the graph also I obviously need to brush up on my trig functions!
0
1 week ago
#5
(Original post by id0ntkn0w)
Yes, thats it. Im afraid I dont quite understand why it has stationary points at (2k+1)pi?
Also, in point 7, i dont understand after it says "Also for all integers K". Ive included a photo of the graph also I obviously need to brush up on my trig functions!
You can see where the gradient is "flat" at pi, -pi, 3pi, -3pi, 5pi, -5pi ... It doesn't look totally flat (stationary points) but it is. That is where the "all integers k" comes into it. The original function is
x + sin(x)
and its derivative is
1 + cos(x)
it must be flat whenever cos(x) is -1 and this is x = pi + k2pi. Otherwise the derivative is postive and the graph is monotonic increasing.

Another way of looking at it is that the gradient of x is 1, so the gradient of sin(x) must be -1 at these points. If you rotate your graph 45 degrees clockwise, you can see the familiar trig function and eyeball why this occurs Your trig function never stops and repeats every 2pi (360 degrees), hence the +k2pi condition.
Last edited by mqb2766; 1 week ago
0
#6
(Original post by dextrous63)
The 2k+1 bit refers to every odd multiple of pi.
I had a suspicion it might be that. I just dont understand why, or what impact that has in point 7 after it says "Also, for all integers k". I feel I must have overlooked some of my trig functions study last year! Im good with rational and quadratic but trig obviously needs some work. Could anyone recommend a site or video that might help. Feeling a bit silly
0
1 week ago
#7
(Original post by id0ntkn0w)
I had a suspicion it might be that. I just dont understand why, or what impact that has in point 7 after it says "Also, for all integers k". I feel I must have overlooked some of my trig functions study last year! Im good with rational and quadratic but trig obviously needs some work. Could anyone recommend a site or video that might help. Feeling a bit silly
Nothing to feel silly about. cosx = 1 for every even multiple of pi, and -1 for every odd multiple. Sketch it over a few cycles (both left and right of the y-axis) to convince yourself. 2k+1 generates the sequence ....-5, -3, -1, 1, 3, 5, 7,...., and the "all integers" refers to the fact that the cos curve travels left as well as right of the y-axis.

It's good that you had an instinct/suspicion. The trick is to try and work out if your suspicions are right, and indeed why.
0
#8
(Original post by dextrous63)
Nothing to feel silly about. cosx = 1 for every even multiple of pi, and -1 for every odd multiple. Sketch it over a few cycles (both left and right of the y-axis) to convince yourself. 2k+1 generates the sequence ....-5, -3, -1, 1, 3, 5, 7,...., and the "all integers" refers to the fact that the cos curve travels left as well as right of the y-axis.

It's good that you had an instinct/suspicion. The trick is to try and work out if your suspicions are right, and indeed why.
Thank you so much, that really makes sense
0
#9
(Original post by mqb2766)
You can see where the gradient is "flat" at pi, -pi, 3pi, -3pi, 5pi, -5pi ... It doesn't look totally flat (stationary points) but it is. That is where the "all integers k" comes into it. The original function is
x + sin(x)
and its derivative is
1 + cos(x)
it must be flat whenever cos(x) is -1 and this is x = pi + k2pi. Otherwise the derivative is postive and the graph is monotonic increasing.

Another way of looking at it is that the gradient of x is 1, so the gradient of sin(x) must be -1 at these points. If you rotate your graph 45 degrees clockwise, you can see the familiar trig function and eyeball why this occurs Your trig function never stops and repeats every 2pi (360 degrees), hence the +k2pi condition.
Thank you so much for your detailed reply. I totally get it now
0
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