x Turn on thread page Beta
 You are Here: Home >< Maths

1. I always create new threads when im stuck on an A level ques. So from now on ill just post here - and people can help. First ques is about CAST circles with radians from C2 - does anyone know about cast circles?
2. Many people on here will know about CAST diagrams, You're going to have to be a bit more specific.
3. (Original post by kiiten)
I always create new threads when im stuck on an A level ques. So from now on ill just post here - and people can help. First ques is about CAST circles with radians from C2 - does anyone know about cast circles?
Watching this is very helpful, I find.
4. (Original post by B_9710)
Many people on here will know about CAST diagrams, You're going to have to be a bit more specific.
Wait ill post the question and my working
5. (Original post by kiiten)
Wait ill post the question and my working
6. Question 2.k)
So i started from pi/2 to get 1/6pi as a solution. Is this wrong? NOTE: i drew the angle from 0 and when i start from 0 i get the right answer? (Sorry if that doesnt make sense)

Posted from TSR Mobile
Attached Images

7. (Original post by kiiten)
Question 2.k)
So i started from pi/2 to get 1/6pi as a solution. Is this wrong? NOTE: i drew the angle from 0 and when i start from 0 i get the right answer? (Sorry if that doesnt make sense)

Posted from TSR Mobile
Remember your CAST diagram rules? You should be doing but then your "key-angle" is and not . Then because is negative, your solutions will be in the second and third quadrant. So and...? You tell me.
8. (Original post by Zacken)
Remember your CAST diagram rules? You should be doing but then your "key-angle" is and not . Then because is negative, your solutions will be in the second and third quadrant. So and...? You tell me.
I'm confused

cos (x + pi/2) = -1/2
let u = x + pi/2

so cosu = -1/2
u = 2/3 pi

is that right so far? - I don't understand what you mean after you mention "key-angle"
9. (Original post by kiiten)
so cosu = -1/2
u = 2/3 pi
This bit is not quite correct. It should be:

So u = pi + cos^(-1) (1/2) and u = pi - cos^(-1) (1/2).

Remember when you're doing CAST diagrams, you take the inverse trig of modulus of the RHS.
10. [QUOTE=Zacken;63278041]This bit is not quite correct. It should be:

So u = pi + cos^(-1) (1/2)

where did the pi come from and where did the minus from the 1/2 go?
11. (Original post by kiiten)

where did the pi come from and where did the minus from the 1/2 go?
Did you watch the video that linked you? That should explain everything.
12. (Original post by Zacken)
Did you watch the video that linked you? That should explain everything.
yeah ive watched the video but it only seems to explain the basics. E.g. when I do the inverse of cos (-1/2) = 2/3 pi but you said something about adding pi ?
13. (Original post by kiiten)
yeah ive watched the video but it only seems to explain the basics. E.g. when I do the inverse of cos (-1/2) = 2/3 pi but you said something about adding pi ?
Let me run through an example for you. . Now I'm going to use something called my key angle, which is .

Now, I also know that is negative, so my solutions will be in the second and fourth quadrant. Which means I need to do: and .

If it was , my key-angle would still be but this time is positive, so my solutions will be in the third and first quadrant, so my solutions are and .

You getting the gist of the method?
14. A little bit - i havent learnt this way before. So for the key angle do you always use that sin-1 (1/2) = 30?
15. (Original post by kiiten)
A little bit - i havent learnt this way before. So for the key angle do you always use that sin-1 (1/2) = 30?
You should know that sin-1 (1/2) = 30. This is because sin(30) = 1/2.
16. (Original post by Zacken)
You should know that sin-1 (1/2) = 30. This is because sin(30) = 1/2.
I asked my friend for help - i think i know where i went wrong. If you look at the diagram i attached earlier i was calculating the solution from the bound pi/2 rather than from 0 so my answer would be slightly less. I.e. the angle started from 0 and stopped after pi/2 so you still use the full angle. If that makes any sense?

Posted from TSR Mobile
17. General cast circle ques

E.g. if the bounds are from pi/2 to 5/2 pi and the angle is 2/3 pi do you calculate the solution from 0 (green arrow) or pi/2 (red arrow)?

Posted from TSR Mobile
Attached Images

18. (Original post by kiiten)
General cast circle ques

E.g. if the bounds are from pi/2 to 5/2 pi and the angle is 2/3 pi do you calculate the solution from 0 (green arrow) or pi/2 (red arrow)?

Posted from TSR Mobile
If the bounds are between pi/2 and 5pi/2 then it makes sense to only look for solutions between pi/2 and 5pi/2. So you would look for solutions from the red arrow.

But angles will still start from 0.

Your question isn't completely clear so it may help if you post an example question if you're still not sure.
So the red arrow.
19. (Original post by notnek)
If the bounds are between pi/2 and 5pi/2 then it makes sense to only look for solutions between pi/2 and 5pi/2. So you would look for solutions from the red arrow.

But angles will still start from 0.

Your question isn't completely clear so it may help if you post an example question if you're still not sure.
So the red arrow.
If the angle is 2/3 pi and i use the red arrow the answer is wrong - i get 1/6 pi. But if you said the angle starts from 0 shouldnt it be from the green arrow because the angle stops after the bound?

Posted from TSR Mobile
20. (Original post by kiiten)
If the angle is 2/3 pi and i use the red arrow the answer is wrong - i get 1/6 pi. But if you said the angle starts from 0 shouldnt it be from the green arrow because the angle stops after the bound?

Posted from TSR Mobile
yes always measure angles from the green arrow

what notnek is saying is that solutions between from where the red arrow starts until a full rotation later (ie 1.25 cirlces) are the only valid ones

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 29, 2016
Today on TSR

### Any tips for freshers...

who are introverts?

### More snow?!

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE