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Trigonometric proofs

Me and my friends are all completely stuck on this if anyone could give a few pointers

Prove that sin3A= 2cosAsin2A- (1/2cosA)sin2A

So far I've tried using the addition formula saying that
sin3A= sin(A+2A) = sinAcos2A + cosAsin2A
I've used double angle formulea and subsituted for cos2A and sin2A but nothing I'm doing seems to work and I just get myself back to where I started or to sin3A= 3sinAcos^2A- sin^3 A

So yeah if anyone has any ideas or suggestions I'd be really grateful :smile:

Reply 1

ghostwalker

Original post by Elinh

So yeah if anyone has any ideas or suggestions I'd be really grateful :smile:

Reduce both sides to terms in SinA and CosA.

Edit: What you end up with is much easier to prove, IMO.

(edited 7 years ago)

Reply 2

DFranklin

Original post by ghostwalker

Reduce both sides to terms in SinA and CosA.

Edit: What you end up with is much easier to prove, IMO.

Helps if you know the formula for sin3x in terms of sin x...

Reply 3

ghostwalker

Original post by DFranklin

Helps if you know the formula for sin3x in terms of sin x...

I could never remember the triple angle formulae - so rarely used, for my part - and always had to work them from scratch.

Reply 4

DFranklin

Original post by ghostwalker

I could never remember the triple angle formulae - so rarely used, for my part - and always had to work them from scratch.

Me either, but I was lazy enough to google it. Was really just giving an extra hint to OP...

Attachment not found

Original post by Anfanny

Similar to factor theorum you use the actual identities to make more identities that you can substitute in to the original equation. By equating the correct terms. When you do it one way it does no work and it should be faster than the above method.There are around 10 equality/equivalence signs in what you've posted, and other than the initial idenity involving sin(A+2A), I don't think any of them are correct.

e.g.

sin 2A = 1/2 sin 2A ???

2 cos A = cos A ???

etc.

(edited 7 years ago)

Reply 7

username2769500

Original post by DFranklin

There are around 10 equality/equivalence signs in what you've posted, and other than the initial idenity involving sin(A+2A), I don't think any of them are correct.

e.g.

sin 2A = 1/2 sin 2A ???

2 cos A = cos A ???

etc.

Mine was trial and error. I will do it again in pen. It should be equating signs as it's an identity and not a solution for this part of the question it looks awkward but equal signs mean nothing when doing this as I am equating terms and making more identities once the correct parts are equated from both equations after sin(A+2A) is expanded.

(edited 7 years ago)

Reply 8

Elinh

Ok thanks everyone for the advice I've got it now

Reply 9

DFranklin

The point is that everything you are marking as "equal" has to be true for all A (since you're looking for an idenity).

Is it true that cos 2A = 2 cos A (for all values of A)? No. So any part of working that shows this, claims this, or uses this, is invalid.

Edit: above reply is for Anfanny...

Reply 10

DFranklin

Original post by Elinh

Ok thanks everyone for the advice I've got it now

Great. Small bit of advice: as a rule of thumb, with problems like this it's usually easier to start with the more complicated side, as unintuitive as it might seem. (It's not like it's 100% always the best, more like 60:40 it's the way to go...)

Reply 11

username2769500

Original post by DFranklin

Okay I thought you were asking for a better proof. This way is correct and very fast. It has to work because there's more way than one to get the correct answer and when I equate the wrong terms it does not make itself up to the identity on the right so it is correct. Please write me out a proof saying it is wrong.

Reply 12

username2769500

Original post by Elinh

Ok thanks everyone for the advice I've got it now

That's good! Usually just expanding stuff out simply you will get the right answer I'm not 100 percent on it and still need to go over some problems myself.

Reply 13

DFranklin

Original post by Anfanny

Okay I thought you were asking for a better proof. This way is correct and very fast.

No it isn't.

It has to work because there's more way than one to get the correct answer and when I equate the wrong terms it does not make itself up to the identity on the right so it is correct. Please write me out a proof saying it is wrong.

Your proof states that

cos 2A = 2 cos A, and also that cos A = 1 / (2 cos 2A)

Neither of these statements are true.

Therefore your proof is invalid. I suggest you ask someone qualified, who you trust, for their opinion, for I do not intend to spend more time on this, but I can assure you that you will not get a single mark for a "proof" like this in an exam, so it is something you need to sort out.

(edited 7 years ago)

Reply 14

username2769500

Original post by DFranklin

No it isn't.

Your proof states that

cos 2A = 2 cos A, and also that cos A = 1 / (2 cos 2A)

Neither of these statements are true.

Therefore your proof is invalid. I suggest you ask someone qualified, who you trust, for their opinion, for I do not intend to spend more time on this, but I can assure you that you will not get a single mark for a "proof" like this in an exam, so it is something you need to sort out.

I want to see a piece of paper where you have an equal sign with a diagonal line through it. It is correct. The other one is negative not posotive. The function was not defined in this question. My way is correct.

Reply 15

Mr M

Study Forum Helper

Original post by Anfanny

...

In case you are interested in a second opinion, I can confirm that there are many errors in the answer you posted.

Reply 16

username2769500

Can you show me where they are? As I have spent a lot of effort on my method and I came up with this answer which uses only making new identities, is quick and does not require a lot of work.

(edited 7 years ago)

Reply 17

Mr M

Study Forum Helper

Original post by Anfanny

Can you show me where they are? As I have spent a lot of effort on my method and I came up with this answer which uses only making new identities, is quick and does not require a lot of work.

DFranklin has already pointed out four identities that are completely wrong.

Here's one example:

2 \cos A = \cos A

This statement is only true when

\cos A=0

. It needs to be true for all values of A.

Reply 18

username2769500

It should be multiplied by Something else you have to look closely in my 3rd image as I used trial and error to come up with my answer. It might have been cos(2A) = 2cosA or 2cosAcosAsinA =2cosAsin(2A)