Jshek
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Dont have a clue on 7d and 8 no example on youtube tells me how to
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Jshek
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LiberOfLondon
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π radians = 180 degrees
2π radians = 360 degrees
1 radian ≈ 57.324 degrees
1 degree ≈ 0.017 radians

Hope this helps. Don't blame me for the weird units, radians are somehow more irritating to work with then the imperial system.
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Jshek
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(Original post by LiberOfLondon)
π radians = 180 degrees
2π radians = 360 degrees
1 radian ≈ 57.324 degrees
1 degree ≈ 0.017 radians

Hope this helps. Don't blame me for the weird units, radians are somehow more irritating to work with then the imperial system.
What am I suppose to do with that
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LiberOfLondon
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(Original post by Jshek)
What am I suppose to do with that
They're conversion tables between radians and degrees.
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zetamcfc
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(Original post by LiberOfLondon)
They're conversion tables between radians and degrees.
I don't think OP has a problem with that, rather just the specific questions they have asked...
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davros
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(Original post by Jshek)
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For 7d, could you plot y = sin(x/3) if required? The "+1" at the end is just going to be a vertical shift applied to this...
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Jshek
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(Original post by davros)
For 7d, could you plot y = sin(x/3) if required? The "+1" at the end is just going to be a vertical shift applied to this...
The answer attached I dont get it
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davros
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(Original post by Jshek)
The answer attached I dont get it
Where is your confusion? They haven't gone all the way up to 6\pi unless you've cut off the image, but it should be straightforward.
You know that sin 0 = 0 so (sin 0 ) + 1 = 1 which gives you the intersection on the y-axis.

Also the first time that sin x reaches its minimum (-1) is at x = 3\pi/2 so if you're looking for the corresponding point on the graph of y = sin(x/3) it will be at x = 9\pi/2. Then y = sin(x/3) + 1 = -1 + 1 = 0 so this is the first intersection with the (positive) x-axis.
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LiberOfLondon
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(Original post by zetamcfc)
I don't think OP has a problem with that, rather just the specific questions they have asked...
Ah. I misread the OP then.
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Jshek
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(Original post by davros)
Where is your confusion? They haven't gone all the way up to 6\pi unless you've cut off the image, but it should be straightforward.
You know that sin 0 = 0 so (sin 0 ) + 1 = 1 which gives you the intersection on the y-axis.

Also the first time that sin x reaches its minimum (-1) is at x = 3\pi/2 so if you're looking for the corresponding point on the graph of y = sin(x/3) it will be at x = 9\pi/2. Then y = sin(x/3) + 1 = -1 + 1 = 0 so this is the first intersection with the (positive) x-axis.
Thanks I get it now but how do I do question 8?
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davros
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(Original post by Jshek)
Thanks I get it now but how do I do question 8?
So for the x-axis intercepts you need to be solving \cos (x - \dfrac{2\pi}{3}) = 0 in the required region (domain of x values).

Start by thinking about when cos w = 0 for some angle w.
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Jshek
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(Original post by davros)
So for the x-axis intercepts you need to be solving \cos (x - \dfrac{2\pi}{3}) = 0 in the required region (domain of x values).

Start by thinking about when cos w = 0 for some angle w.
I'm really busy can u just tell me thanks
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davros
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(Original post by Jshek)
I'm really busy can u just tell me thanks
Well I'm busy too actually

I'm not going to tell you the answer - that's not how this forum works! You should be able to work this out by looking at the other examples you've worked on - it's basically a sideways translation of the standard cos graph.
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