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# Official Edexcel C3 Discussion Thread watch

1. I don't know anything about arccos, and arcsin and all that. What do we need to know about them?
2. (Original post by placenta medicae talpae)
I find the double one for cos easy to remember because it's like the cos-sin formula for 1, but with the wrong sign.
So instead of being , it's .

And then I remember the one for as the one with the 2 in front
I just...remember them. I'm pretty sure my mind has some method for the memory, but I won't know, I don't visit my mind very often!
3. Ya if you want to turn dx/dy into dy/dx just stick dx/dy over 1 :')

dy/dx = 1/(dx/dy)
4. (Original post by Inzamam99)
could some one plz help me differentiate: x= 2cos(2y+pi)

I think...
5. (Original post by jimber)
Thanks, I prey for none of these q's on the exam!
they're actually quite easy!

(Original post by I Have No Imagination)
I don't know anything about arccos, and arcsin and all that. What do we need to know about them?
arccos and arcsin is the same as cos-1 and sin-1 basically the inverse of sine and cosine etc.
And yes, we need to know about them they're part of the trig chapters!
6. (Original post by Doughboy)

I think...
that's right!
7. A bit of light revision on graphs like e^x and lnx tonight i think. And i have to teach myself how to "complete the square" as i missed them lessons >.< Anyone got the formula for completing the square by any chance, and a quick example of how to use it?

Then some more revision tomorrow morning if i can get the lessons off.
8. (Original post by unamed)
they're actually quite easy!

arccos and arcsin is the same as cos-1 and sin-1 basically the inverse of sine and cosine etc.
And yes, we need to know about them they're part of the trig chapters!
We also need to know their graphs! They the inverse trig functions, so all we'd need to do it reflect the original trig graphs in line y=x to get those inverse graphs.

Anyway, I've seen Edexcel tie in the inverse trig with differentiation:

9. (Original post by Doughboy)
We also need to know their graphs! They the inverse trig functions, so all we'd need to do it reflect the original trig graphs in line y=x to get those inverse graphs.

Anyway, I've seen Edexcel tie in the inverse trig with differentiation:

Or you could memorise the graphs (< I recommend this)
that's a not-nice question. Which paper is it in? {where did you see it?}
10. (Original post by Doughboy)

I think...
Thank you and thanks to unamed as well
11. How would you do part B ?

Thanks
12. Eek, exam tomorrow. I hope it won't be too bad, I need to get 90% in it!! (:
Good luck to everyone taking it (:
13. (Original post by racshot65)
How would you do part B ?

Thanks
I can't be bothered to do latex (sorry)
3sin2x+4cos2x=Rsin2xcosa + Rcos2xsina

R2=32+42
therefore R=5

and 3=cosa and 4=sina
therefore tana=4/3
and you can work it out from there!
14. This thread is going to have to shut down tomorrow right?
15. fingers crossed its going to be easy
16. (Original post by racshot65)
How would you do part B ?

Thanks
Well you expand Rsin(2x+alpha)

= Rsin2x.cos(alpha) + Rcos2x.sin(aplha)

then compare the co-efficiants to that of the orignal

so you get

3sin2x + 4cos2x = Rsin2x.cos(aplha) + Rcos2x.sin(alpha)

3 = Rcos(alpha) (1)

4 = Rsin(alpha) (2)

(2)/(1) = tan(alpha) = 4/3

(alpha) = 53.13 (2d.p)

(1)^2 + (2)^2 = R^2 = 25

R=5

Hope that helps
17. (Original post by Snapshot13)
A bit of light revision on graphs like e^x and lnx tonight i think. And i have to teach myself how to "complete the square" as i missed them lessons >.< Anyone got the formula for completing the square by any chance, and a quick example of how to use it?

Then some more revision tomorrow morning if i can get the lessons off.
My teacher at GCSE gave me these simple steps:

1) Separate the 'c' term from the 'a' and 'b' terms. Factor out the co-efficient of to form an expression with brackets.

2) Then add the SQUARE of HALF the CO-EFFICIENT OF X to the brackets. Do not simplify this square!

3) Then subtract the SQUARE of HALF the CO-EFFICIENT OF X outside the brackets. Simplify this square.

4) Delete the middle term within the brackets to form your perfect square.
I'll derive the quadratic formula by applying these steps:

1) Factor out the co-efficient of to form an expression with brackets.

2) Then add the SQUARE of HALF the CO-EFFICIENT OF X to the brackets. Do not simplify this square!

Half the co-efficient of x is: . The square of this is:

Adding this to the brackets:

3) Then subtract the SQUARE of HALF the CO-EFFICIENT OF X outside the brackets. Simplify this square.

4) Delete the middle term within the brackets and form your perfect square.

If you equate this to zero, you will end up with the quadratic formula.

Hope this wasn't too confusing.
18. (Original post by Doughboy)
My teacher at GCSE gave me these simple steps:

I'll derive the quadratic formula by applying these steps:

1) Factor out the co-efficient of to form an expression with brackets. Take out the third term from the brackets.

=

2) Then add the SQUARE of HALF the CO-EFFICIENT OF X to the brackets. Do not simplify this square!

Half the co-efficient of x is: . The square of this is:

Adding this to the brackets:

3) Then subtract the SQUARE of HALF the CO-EFFICIENT OF X outside the brackets. Simplify this square.

4) Delete the middle term within the brackets and form your perfect square.

If you equate this to zero, you will end up with the quadratic formula.

Hope this wasn't too confusing.
that wasn't very simple

all you have to do is make sure the coefficient of x^2 is 1
19. (Original post by ahnaf.c)
This thread is going to have to shut down tomorrow right?
i guess so.
20. No it wasnt simple :P I prefer the other formula pheylan just gave

So what can "completing the square" be used for? just thinking of its practical use =/

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