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# C3 Factor Formulae/R Formulae? [Edexcel] watch

1. Are students doing the edexcel c3 exam required to know the Factor and R formulae?

Thank you
2. (Original post by jamb97)
Are students doing the edexcel c3 exam required to know the Factor and R formulae?

Thank you
Factor - no.

R - yes.
3. (Original post by Zacken)
Factor - no.

R - yes.
Hi could you tell me how I would simplify sin^2 theta x cos^2 theta using the double angle formulas?
4. (Original post by mathshelp1956)
Hi could you tell me how I would simplify sin^2 theta x cos^2 theta using the double angle formulas?

But you know that

So .

I'm sure you can take it from here.
5. (Original post by Zacken)

But you know that

So .

I'm sure you can take it from here.
Thank you so much.
6. (Original post by mathshelp1956)
Thank you so much.
No worries.
7. (Original post by Zacken)
No worries.
Hi, I'm sorry to bother you again. Im stuck on simplifying 4sintheta costheta cos2theta. I just don't know where to start. Again, thanks for the help.
8. (Original post by mathshelp1956)
Hi, I'm sorry to bother you again. Im stuck on simplifying 4sintheta costheta cos2theta. I just don't know where to start. Again, thanks for the help.
what?
9. (Original post by Zacken)
what?
I don't understand how you went to the second line of 2cos2theta X 2sinthetacostheta
10. (Original post by mathshelp1956)
I don't understand how you went to the second line of 2cos2theta X 2sinthetacostheta
.

So . But the latter bit is just .

Hence
11. I understand the last line. But how would you go from 2 X Sin2thetaCos2theta to Sin4theta?
(Original post by Zacken)
.

So . But the latter bit is just .

Hence
12. (Original post by mathshelp1956)
I understand the last line. But how would you go from 2 X Sin2thetaCos2theta to Sin4theta?
.

Here so .

Basically the double angle tells you 2 sin(anything) cos(anything) = sin (2 * anything), here your anything is and 2*anything is
13. (Original post by Zacken)
.

Here so .

Basically the double angle tells you 2 sin(anything) cos(anything) = sin (2 * anything), here your anything is and 2*anything is
I can't thank you enough. By the way, would this be considered a more difficult trigonometry question?
14. (Original post by mathshelp1956)
I can't thank you enough. By the way, would this be considered a more difficult trigonometry question?
Don't worry about it. Uhm, I don't think so. I can see why it'd seem like that, but honestly, you'll get used to all these manipulations very fast with practice.
15. (Original post by Zacken)
Factor - no.

R - yes.
Thanks for your answer! I want to give you rep but the forum will not let me. Could you also help me with this proving identity question please?

Prove that:

1/(cosecx+cotx) is identical to 1-cosx/sinx.

I have got up to sinx/1+cosx by manipulating the right hand side after not finding the left hand side doable, but I don't think I'm getting anywhere..
16. (Original post by jamb97)
Thanks for your answer! I want to give you rep but the forum will not let me. Could you also help me with this proving identity question please?

Prove that:

1/(cosecx+cotx) is identical to 1-cosx/sinx.

I have got up to sinx/1+cosx by manipulating the right hand side after not finding the left hand side doable, but I don't think I'm getting anywhere..
If I tell you that:

Can you take it from here?

This is a bit tough to spot, I must admit.
17. (Original post by Zacken)
If I tell you that:

Can you take it from here?

This is a bit tough to spot, I must admit.
Yeah I got it thanks, converted the denominator to sin^(2)x and it simplified nicely. You seem to be quite good at A-level maths so I hope you won't mind if I ask for your help again please, this time with a M1 question.

This is the M1 Delphis paper 4, question 2 and can be found here: http://www.physicsandmathstutor.com/...rs/m1-delphis/

My method was to resolve the 4N and 5N forces parallel to the 8N force. So I did 8N - (4cos30+5sin60) = 0.2... which is not correct. Could you explain why my method does not work and how to work out this correctly?

Thank you
18. (Original post by jamb97)
My method was to resolve the 4N and 5N forces parallel to the 8N force. So I did 8N - (4cos30+5sin60) = 0.2... which is not correct. Could you explain why my method does not work and how to work out this correctly?
Why are you resolving in such a weird direction if you want to find the resultant force? Resolve horizontally and vertically, then to get the resultant force, you just square, add and square root.

What you're doing when you do this is forming a force triangle with the base as your horizontal component, height as your vertical component, and your hypotenuse which is is your resultant force. The direction will then be the angle the hypotenuse makes with the horizontal/vertical.

If you resolve parallel to the 8 N force, you'd get the component acting in that direction and not the resultant. So it's of no use in finding the resultant force.
19. (Original post by Zacken)
Why are you resolving in such a weird direction if you want to find the resultant force? Resolve horizontally and vertically, then to get the resultant force, you just square, add and square root.

What you're doing when you do this is forming a force triangle with the base as your horizontal component, height as your vertical component, and your hypotenuse which is is your resultant force. The direction will then be the angle the hypotenuse makes with the horizontal/vertical.

If you resolve parallel to the 8 N force, you'd get the component acting in that direction and not the resultant. So it's of no use in finding the resultant force.
Thanks, I think I understand now. Do you know where this theory comes from in the M1 Edexcel book? It seems a bit unfamiliar, the only thing I can think of is when finding the modulus but I think that's C4
20. (Original post by jamb97)
Thanks, I think I understand now. Do you know where this theory comes from in the M1 Edexcel book? It seems a bit unfamiliar, the only thing I can think of is when finding the modulus but I think that's C4
No, sorry. I've never seen the book.

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