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C3 Factor Formulae/R Formulae? [Edexcel]

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

Thank you

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Reply 1
Avatar for Zacken
Zacken
22
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.
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?
Reply 3
Avatar for Zacken
Zacken
22
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?


sin⁡2θ×cos⁡2θ=sin⁡θsin⁡θcos⁡θcos⁡θ=sin⁡θcos⁡θsin⁡θcos⁡θ\sin^2 \theta \times \cos^2 \theta = \sin \theta \sin \theta \cos \theta \cos \theta = \sin \theta \cos \theta \sin \theta \cos \theta

But you know that sin⁡2θ=2sin⁡θcos⁡θ⇒sin⁡θcos⁡θ=sin⁡2θ2\sin 2\theta = 2 \sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta = \frac{\sin 2\theta}{2}

So sin⁡2θcos⁡2θ=sin⁡2θ2sin⁡2θ2\displaystyle \sin^2 \theta \cos^2 \theta = \frac{\sin 2\theta}{2} \frac{\sin 2\theta}{2}.

I'm sure you can take it from here.
Original post by Zacken
sin⁡2θ×cos⁡2θ=sin⁡θsin⁡θcos⁡θcos⁡θ=sin⁡θcos⁡θsin⁡θcos⁡θ\sin^2 \theta \times \cos^2 \theta = \sin \theta \sin \theta \cos \theta \cos \theta = \sin \theta \cos \theta \sin \theta \cos \theta

But you know that sin⁡2θ=2sin⁡θcos⁡θ⇒sin⁡θcos⁡θ=sin⁡2θ2\sin 2\theta = 2 \sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta = \frac{\sin 2\theta}{2}

So sin⁡2θcos⁡2θ=sin⁡2θ2sin⁡2θ2\displaystyle \sin^2 \theta \cos^2 \theta = \frac{\sin 2\theta}{2} \frac{\sin 2\theta}{2}.

I'm sure you can take it from here.


Thank you so much. :biggrin::biggrin:
Reply 5
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Zacken
22
Original post by mathshelp1956
Thank you so much. :biggrin::biggrin:


No worries.
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.
Reply 7
Avatar for Zacken
Zacken
22
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.


4sin⁡θcos⁡θcos⁡2θ=2cos⁡2θ×2sin⁡θcos⁡θ=2sin⁡2θcos⁡2θ=4 \sin \theta \cos \theta \cos 2\theta = 2\cos 2\theta \times 2\sin \theta \cos \theta = 2\sin 2\theta \cos 2\theta = what?
Original post by Zacken
4sin⁡θcos⁡θcos⁡2θ=2cos⁡2θ×2sin⁡θcos⁡θ=2sin⁡2θcos⁡2θ=4 \sin \theta \cos \theta \cos 2\theta = 2\cos 2\theta \times 2\sin \theta \cos \theta = 2\sin 2\theta \cos 2\theta = what?


I don't understand how you went to the second line of 2cos2theta X 2sinthetacostheta
Reply 9
Avatar for Zacken
Zacken
22
Original post by mathshelp1956
I don't understand how you went to the second line of 2cos2theta X 2sinthetacostheta


4=2×24 = 2\times 2.

So 4cos⁡θsin⁡θ=2×2cos⁡θsin⁡θ4 \cos \theta \sin \theta = 2\times 2 \cos \theta \sin \theta. But the latter bit is just sin⁡2θ=2cos⁡θsin⁡θ\sin 2\theta = 2\cos \theta \sin \theta.

Hence 4cos⁡θsin⁡θcos⁡2θ=2×2cos⁡θsin⁡θ×cos⁡2θ=2×sin⁡2θcos⁡2θ4 \cos \theta \sin \theta \cos 2\theta = 2 \times 2 \cos \theta \sin \theta \times \cos 2\theta = 2 \times \sin 2\theta \cos 2\theta
I understand the last line. But how would you go from 2 X Sin2thetaCos2theta to Sin4theta?
Original post by Zacken
4=2×24 = 2\times 2.

So 4cos⁡θsin⁡θ=2×2cos⁡θsin⁡θ4 \cos \theta \sin \theta = 2\times 2 \cos \theta \sin \theta. But the latter bit is just sin⁡2θ=2cos⁡θsin⁡θ\sin 2\theta = 2\cos \theta \sin \theta.

Hence 4cos⁡θsin⁡θcos⁡2θ=2×2cos⁡θsin⁡θ×cos⁡2θ=2×sin⁡2θcos⁡2θ4 \cos \theta \sin \theta \cos 2\theta = 2 \times 2 \cos \theta \sin \theta \times \cos 2\theta = 2 \times \sin 2\theta \cos 2\theta
Reply 11
Avatar for Zacken
Zacken
22
Original post by mathshelp1956
I understand the last line. But how would you go from 2 X Sin2thetaCos2theta to Sin4theta?


2sin⁡Acos⁡A=sin⁡2A2 \sin A \cos A = \sin 2A.

Here A=2θA = 2\theta so 2sin⁡2θcos⁡2θ=sin⁡(2(2θ))=sin⁡4θ2 \sin 2\theta \cos 2\theta = \sin (2(2\theta)) = \sin 4\theta.

Basically the double angle tells you 2 sin(anything) cos(anything) = sin (2 * anything), here your anything is 2θ2\theta and 2*anything is 4θ4\theta
Original post by Zacken
2sin⁡Acos⁡A=sin⁡2A2 \sin A \cos A = \sin 2A.

Here A=2θA = 2\theta so 2sin⁡2θcos⁡2θ=sin⁡(2(2θ))=sin⁡4θ2 \sin 2\theta \cos 2\theta = \sin (2(2\theta)) = \sin 4\theta.

Basically the double angle tells you 2 sin(anything) cos(anything) = sin (2 * anything), here your anything is 2θ2\theta and 2*anything is 4θ4\theta


I can't thank you enough. By the way, would this be considered a more difficult trigonometry question?
Reply 13
Avatar for Zacken
Zacken
22
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. :yes:
Reply 14
Avatar for jamb97
jamb97
OP
2
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..
Reply 15
Avatar for Zacken
Zacken
22
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:

sin⁡x1+cos⁡x×1−cos⁡x1−cos⁡x\displaystyle \frac{\sin x}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x}

Can you take it from here?

This is a bit tough to spot, I must admit.
Reply 16
Avatar for jamb97
jamb97
OP
2
Original post by Zacken
If I tell you that:

sin⁡x1+cos⁡x×1−cos⁡x1−cos⁡x\displaystyle \frac{\sin x}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x}

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/a-level-maths-papers/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 :smile:
Reply 17
Avatar for Zacken
Zacken
22
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 a2+b2\sqrt{a^2 + b^2} 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.
Reply 18
Avatar for jamb97
jamb97
OP
2
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 a2+b2\sqrt{a^2 + b^2} 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
Reply 19
Avatar for Zacken
Zacken
22
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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