Radians

Jshek

Dont have a clue on 7d and 8 no example on youtube tells me how to

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.

Reply 3

Jshek

Original post by LiberOfLondon

What am I suppose to do with that

Reply 4

username4406682

Original post by Jshek

What am I suppose to do with that

They're conversion tables between radians and degrees.

Reply 5

zetamcfc

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...

Reply 6

davros

Original post by Jshek

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...

Reply 7

Jshek

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

Screenshot_20200630-162642_Word.jpg236.2KB

Reply 8

davros

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.

Reply 9

username4406682

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.

Reply 10

Jshek

Original post by davros

Where is your confusion? They haven't gone all the way up to

6\pi

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?

Reply 11

davros

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.

Reply 12

Jshek

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

Reply 13

davros

Original post by Jshek

I'm really busy can u just tell me thanks

Well I'm busy too actually :smile:

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.