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Trigonometric proofs Watch

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    Name:  20170110_145228.jpg
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Size:  516.1 KBMe 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
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    (Original post by Elinh)
    So yeah if anyone has any ideas or suggestions I'd be really grateful
    Reduce both sides to terms in SinA and CosA.

    Edit: What you end up with is much easier to prove, IMO.
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    (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...
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    (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.
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    (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...
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    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.
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    (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.
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    (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.
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    Name:  WIN_20170110_16_12_08_Pro.jpg
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Size:  99.5 KBOk thanks everyone for the advice I've got it now
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    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...
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    (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...)
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    (Original post by 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...
    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.
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    (Original post by Elinh)
    Name:  WIN_20170110_16_12_08_Pro.jpg
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    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.
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    (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.
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    (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.
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    (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.
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    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.
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    (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.
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    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)
 
 
 
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