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    Can someone please explain to me how to use the double angle formulae to work these out?
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    (Original post by futureaussiecto)
    Can someone please explain to me how to use the double angle formulae to work these out?
    *gets out formulae booklet*


    ok, so the single angle formula is;

    sin (A + B) = sin A cos B + sin B cos A

    which is in the formula book.

    now ok, the double angle formula [sin(2x)] is basically;

    sin (x + x) = sin x cos x + sin x cos x

    which is also

    sin (2x) = 2sin x cos x

    so, if we were to look at

    sin(3x)

    this is basically;

    sin(1.5x + 1.5x) = ...

    so do the subbing in into the single angle formula;

    2 sin (1.5x) cos (1.5x)

    so therefore;

    Sin(3x) = 2 sin 1.5x cos 1.5x


    checking values;

    let x=2

    sin(3*2) = 2(sin (1.5*2) * cos (1.5*2)

    which is right
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    see i understand that

    but in my book it says

    Sin3A

    =sin(2A+A)
    =Sin2AcosA+cos2ASinA

    which threw me
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    No you have to find it in terms of sinx.
    Basically, use the double angle formulae to express sin3x in terms of sin2x and sinx, then use the double angle formulae again to get the sin2x down to sinx.
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    (Original post by futureaussiecto)
    see i understand that

    but in my book it says

    Sin3A

    =sin(2A+A)
    =Sin2AcosA+cos2ASinA

    which threw me
    Yes that is true.
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    TSR Group Staff
    That formula from your book is just using the addition formulae instead of the double angle ones, perfectly acceptable, and a bit easier for odd numbers I guess. A and B are just numbers, if you get 5A just let it be C or something and carry on. Nothing sacred about letters.
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    (Original post by futureaussiecto)
    Can someone please explain to me how to use the double angle formulae to work these out?
    sin(2x)=sin(x+x)=sinxcosx+cosxsi nx=2sinxcosx
    cos(2x)=cos(x+x)=cosxcosx-sinxsinx=cos^2x-sin^2x
    sin(3x)=sin(2x+x)=sin2xcosx+cos2 xsinx,,,now use expressions for cos(2x) and sin(2x)
    cos(3x)=cos(2x+x) etc
    sin(4x)=sin(2x+2x) or sin(3x+x) etc
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    (Original post by futureaussiecto)
    see i understand that

    but in my book it says

    Sin3A

    =sin(2A+A)
    =Sin2AcosA+cos2ASinA

    which threw me
    ahh


    ok, i did this;

    sin (1.5x + 1.5x) = ...

    all they did was say

    sin (2x + x) = ...

    and worked it out like that instead.

    in both cases, the 3x was split up into two parts, but they were split differently.
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    The idea is you break it down so there's only terms involving sinx or cosx, no sin2x or sin3x etc. So just use the addition formulae to get it in terms of sin2x and cos2x, and the double-angle formulae to break it down further. It just involves a lot of substitution.
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    so sin(1x) is just sin x?

    if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??
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    (Original post by futureaussiecto)
    so sin(1x) is just sin x?
    Yes/
    if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??
    No/
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    (Original post by futureaussiecto)
    if so, isnt sin(3x) just sin(2x) [which is 2sinxcosx} + sinx??
    No:

    sin(3x)
    = sin (2x + x)
    = sin2x.cosx + cos2x.sinx
    = 2sinx.cosx.cosx + (1-2sin2x)sinx
    = 2sinx.cos2x + sinx - 2sin3x
    = 2sinx(1-sin2x) + sinx - 2sin3x
    = 2sinx - 2sin3x + sinx - 2sin3x
    = 3sinx - 4sin3x
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    (Original post by Libertine)
    No:

    sin(3x)
    = sin (2x + x)
    = sin2x.cosx + cos2x.sinx
    ok from that it looks like

    sin2x = sin2xcosx
    and
    sinx = cos2xsinx

    :eek:
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    (Original post by futureaussiecto)
    ok from that it looks like

    sin2x = sin2xcosx
    and
    sinx = cos2xsinx

    :eek:
    sin (2x + x) does not expand to sin2x + sinx

    Never do that with trig functions.
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    (Original post by Dekota)
    sin (2x + x) does not expand to sin2x + sinx

    Never do that with trig functions.
    yeh i know, i was just saying from the prev quote that when it said

    sin(2x + x) = sin2x.cosx + cos2x.sinx

    that (2x) is the sin2x.cosx bit

    and

    (x) is the cos2x.sinx
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    (Original post by futureaussiecto)
    yeh i know, i was just saying from the prev quote that when it said

    sin(2x + x) = sin2x.cosx + cos2x.sinx

    that (2x) is the sin2x.cosx bit

    and

    (x) is the cos2x.sinx
    But it's not. The whole expression on the right is equal to the whole expression on the right. The bits before and after the addition signs don't correspond.
 
 
 
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