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sin(4theta) trig identity proof needed please

Would it be possible for someone to possibly write out a brief proof for this jump? It's a bit of a large jump for me to be able to see exactly how they moved from this stage to the next. I've tried to prove it, and failed! Maybe I'm just missing something?

Thanks
proof.JPG5.7KB

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Original post by Mutleybm1996
Would it be possible for someone to possibly write out a brief proof for this jump? It's a bit of a large jump for me to be able to see exactly how they moved from this stage to the next. I've tried to prove it, and failed! Maybe I'm just missing something?

Thanks


sin⁡4ϕ=sin⁡2(2ϕ)=2sin⁡(2ϕ)cos⁡(2ϕ)=⋯\sin 4\phi = \sin 2(2\phi) = 2\sin(2\phi)\cos(2\phi) = \cdots

Can you do the rest?
Original post by atsruser
sin⁡4ϕ=sin⁡2(2ϕ)=2sin⁡(2ϕ)cos⁡(2ϕ)=⋯\sin 4\phi = \sin 2(2\phi) = 2\sin(2\phi)\cos(2\phi) = \cdots

Can you do the rest?


Gottit!

thanks :smile:

also, with this one, why don't they do cos(k+1)theta -sin(k+1)theta +i...

unless I'm missing something?
2MS.JPG19.7KB
2Q.JPG12.6KB
Original post by atsruser
sin⁡4ϕ=sin⁡2(2ϕ)=2sin⁡(2ϕ)cos⁡(2ϕ)=⋯\sin 4\phi = \sin 2(2\phi) = 2\sin(2\phi)\cos(2\phi) = \cdots

Can you do the rest?


Like this?
to check.JPG65.1KB
Reply 4
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TeeEm
19
give me 15 minutes please
(edited 8 years ago)
Reply 5
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TeeEm
19
Original post by Mutleybm1996
Like this?


which bit do you need help ... point the post

It looks atsruser is helping you.
Original post by Mutleybm1996
Gottit!

thanks :smile:

also, with this one, why don't they do cos(k+1)theta -sin(k+1)theta +i...

unless I'm missing something?


You can't assume the answer.
Original post by Mutleybm1996
Gottit!

thanks :smile:

also, with this one, why don't they do cos(k+1)theta -sin(k+1)theta +i...

unless I'm missing something?


Original post by Mutleybm1996
Like this?


Original post by TeeEm
which bit do you need help ... point the post

It looks atsruser is helping you.


These posts

And they've gone offline, this needs to be in for tomorrow.
Original post by keromedic
You can't assume the answer.


That was the mark scheme, I'm just trying to see what they do with the bits they don't use in the line below the one I'm referring to.
Reply 9
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TeeEm
19
Original post by Mutleybm1996
These posts

And they've gone offline, this needs to be in for tomorrow.


in which post is the question?
Reply 10
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TeeEm
19
if it is de Moivre's theorem by induction look at this link page 52


http://madasmaths.com/archive/maths_booklets/further_topics/various/proof_by_induction.pdf
Original post by TeeEm
in which post is the question?


sorry, it's two separate questions.
Hold on, I'll quote you in new copies of the question.
Original post by TeeEm
if it is de Moivre's theorem by induction look at this link page 52


http://madasmaths.com/archive/maths_booklets/further_topics/various/proof_by_induction.pdf


Question 1:
is it possible for you to quickly check that the proof(in the shaded boxes) is correct?
to check.JPG65.1KB
Original post by TeeEm
if it is de Moivre's theorem by induction look at this link page 52


http://madasmaths.com/archive/maths_booklets/further_topics/various/proof_by_induction.pdf


Found it(excellent question pack by the way, thank you!)

In the highlighted line, what happens with the bit that disappears(1)?

And the bit underlined as (2), how does this become the isin(k+1)θisin(k+1)\theta ??
(edited 8 years ago)
TeeEm's proof.JPG56.7KB
Reply 14
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TeeEm
19
Original post by Mutleybm1996
Question 1:
is it possible for you to quickly check that the proof(in the shaded boxes) is correct?


assuming phi = arccos(1/root3) this is correct but for a small item which makes no difference.
Original post by TeeEm
assuming phi = arccos(1/root3) this is correct but for a small item which makes no difference.


may i ask what you mean by "for a small item"?

Sorry to bother you...
Reply 16
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TeeEm
19
Original post by Mutleybm1996
Found it(excellent question pack by the way, thank you!)

In the highlighted line, what happens with the bit that disappears(1)?

And the bit underlined as (2), how does this become the isin(k+1)θisin(k+1)\theta ??


this is just the standard expansion of sin(A+B)
sin(A+B) = sinAcosB = cosAsinB


maybe you can see it backwards?

sin(θ+kθ) = sinθcoskθ + cosθsinkθ
(edited 8 years ago)
Reply 17
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TeeEm
19
Original post by Mutleybm1996
may i ask what you mean by "for a small item"?

Sorry to bother you...


sin4θ IS NOT equal to sin22θ in the second grey box first line

I suspect thereis a typo

sin4θ = sin(2 x 2θ
Reply 18
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TeeEm
19
Original post by Mutleybm1996
may i ask what you mean by "for a small item"?

Sorry to bother you...


LOOK AT MY EDIT in post 17 sorry
Original post by TeeEm
this is just the standard expansion of sin(A+B)
sin(A+B) = sinAcosB = cosAsinB


maybe you can see it backwards?

sin(θ+kθ) = sinθcoskθ = cosθsinkθ


OH OH OH!!!!!

Sorry, that was so obvious, I'd just assumed there was a factor being taken out, which left me with two strange unusable terms..well...which I found unusable any way!

Thank you, once again you make me feel like a complete plonker for wasting your time!

Thanks TeeEm :smile:

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