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Math Help - trigo: proving an identity

  1. #1
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    trigo: proving an identity

    Prove the following identities:
    \frac {sin3a}{sina} - \frac {sin3b}{sinb}= 4sin(b+a)sin(b-a)

    Here's what I did:
    L.H.S.
    =3-4sin^2a-(3-4sin^2b)
    =4(sin^2 b - sin^2 a)
    ???

    Thanks in advance.
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  2. #2
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    Hello, cakeboby!

    Prove the following identity: . \frac {\sin3a}{\sin a} - \frac {\sin3b}{\sin b}\:=\: 4\,\sin(b+a)\,\sin(b-a)

    Here's what I did:

    \text{L.H.S.} \:=\:(3-4\sin^2a)-(3-4\sin^2b)

    . . . . . =\;4(\sin^2b - \sin^2a) \qquad\hdots \;\;??
    You're doing great!


    We need some Sum-to-Product idenities:

    . . \begin{array}{ccc}\sin X + \sin Y &=& 2\sin\left(\dfrac{X+Y}{2}\right)\cos\left(\dfrac{X-Y}{2}\right) \\ \\[-3mm]<br />
\sin X - \sin Y &=& 2\cos\left(\dfrac{X+Y}{2}\right)\sin\left(\dfrac{X-Y}{2}\right) \end{array}


    You have: . 4(\sin a + \sin b)(\sin a - \sin b)

    . . . . . . . = \;4\bigg[2\sin\left(\frac{b+a}{2}\right)\cos\left(\frac{b-a}{2}\right)\bigg]\,\bigg[2\cos\left(\frac{b+a}{2}\right)\sin\left(\frac{b-a}{2}\right)\bigg]


    . . . . . . . = \;4\underbrace{\bigg[2\sin\left(\frac{b+a}{2}\right)\cos\left(\frac{b+a  }{2}\right)\bigg]}_{\text{This is }\sin(b+a)}\, \underbrace{\bigg[2\sin\left(\frac{b-a}{2}\right)\cos\left(\frac{b-a}{2}\right)\bigg]}_{\text{This is }\sin(b-a)}


    . . . . . . . =\;4\,\sin(b+a)\,\sin(b-a)

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  3. #3
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    Quote Originally Posted by cakeboby View Post
    Prove the following identities:
    \frac {sin3a}{sina} - \frac {sin3b}{sinb}= 4sin(b+a)sin(b-a)

    Here's what I did:
    L.H.S.
    =3-4sin^2a-(3-4sin^2b)
    =4(sin^2 b - sin^2 a)
    ???

    Thanks in advance.
    you can factorize (sin^2 b - sin^2 a) to continue the equation
    become
    =4(sin (b) - sin (a)) (sin (b) + sin (a))
    and continue with sum to product identities.
    for help use this: http://en.wikipedia.org/wiki/List_of...uct_identities

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  4. #4
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    Hello, cakeboby!


    Do you want to surprise/impress/terrify your professor?


    Prove: . \frac{\sin3a}{\sin a } - \frac{\in3b}{\sin b}\:=\: 4\sin(b+a)\sin(b-a)

    Here's what I did:
    . . LHS\;=\;(3-4\sin^2\!a)-(3-4\sin^2\!b) \;=\;4(\sin^2\! b - \sin^2\! a) ??

    You have: . 4\left[\sin^2\!b-\sin^2\!a\right]


    Subtract and add \sin^2\!a\sin^2\!b:\quad 4\left[\sin^2\!b {\color{red}\:- \:\sin^2\!a\sin^2\!b} \:- \sin^2\!a {\color{red}\:+\: \sin^2\!a\sin^2\!b}\right]


    Factor: . 4\bigg[\sin^2\!b\left(1-\sin^2\!a\right) - \sin^2\!a\left(1-\sin^2\!b\right)\bigg] \;=\; 4\bigg[\sin^2\!b\cos^2\!a - \sin^2\!a\cos^2\!b\bigg]

    Factor: . 4\bigg(\sin b\cos a + \sin a \cos b\bigg)\bigg(\sin b\cos a - \sin a\cos b\bigg) \;=\;4\sin(b+a)\sin(b-a)

    . . . ta-DAA!

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