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Thread: trigo: proving an identity

  1. #1
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    trigo: proving an identity

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

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

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

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

    Here's what I did:

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

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


    We need some Sum-to-Product idenities:

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


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

    . . . . . . .$\displaystyle = \;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] $


    . . . . . . .$\displaystyle = \;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)} $


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

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

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

    Thanks in advance.
    you can factorize $\displaystyle (sin^2 b - sin^2 a)$ to continue the equation
    become
    $\displaystyle =4(sin (b) - sin (a)) (sin (b) + sin (a))$
    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: .$\displaystyle \frac{\sin3a}{\sin a } - \frac{\in3b}{\sin b}\:=\: 4\sin(b+a)\sin(b-a)$

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

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


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


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

    . . . ta-DAA!

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