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Math Help - Proving Trigo Identity

  1. #1
    gp3
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    Proving Trigo Identity

    Hi guys,
    how do you prove these 2:

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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by gp3 View Post
    Hi guys,
    how do you prove these 2:

    LHS

    = \frac{sin3x}{sinx}+\frac{cos3x}{cosx}

    = \frac{sin3xcosx+cos3xsinx}{sinxcosx}

    = \frac{sin(3x+x)}{sinxcosx}

    = \frac{sin4x}{sinxcosx}

    = \frac{2sin2*2x}{2sinxcosx}

    = \frac{2*2sin2xcos2x}{sin2x}

    = 4cos2x

    = RHS

    -----------------------------------------------------------------------------------

    RHS

    = \sqrt{\frac{1}{2}(1-sin\theta)}

    = \sqrt{\frac{1}{2}(cos^2\frac{\theta}{2}+sin^2\frac  {\theta}{2}-2sin\frac{\theta}{2}cos\frac{\theta}{2})}

    = \sqrt{\frac{1}{2}(cos\frac{\theta}{2}-sin\frac{\theta}{2})^2}

    = \frac{1}{\sqrt2}cos\frac{\theta}{2}-\frac{1}{\sqrt2}sin\frac{\theta}{2}

    = sin\frac{\pi}{4}cos\frac{\theta}{2}-cos\frac{\pi}{4}sin\frac{\theta}{2}

    = sin(\frac{\pi}{4}-\frac{\theta}{2})

    = LHS
    Last edited by alexmahone; November 28th 2009 at 10:56 PM.
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  3. #3
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    Hello gp3
    Quote Originally Posted by gp3 View Post
    Hi guys,
    how do you prove these 2:

     \frac{\sin3x}{\sin x}+ \frac{\cos3x}{\cos x}
    =\frac{\sin3x\cos x +\cos 3x \sin x}{\sin x \cos x}

    =\frac{\sin(3x+x)}{\tfrac12\sin2x}

    =\frac{2\sin4x}{\sin2x}

    =\frac{4\sin2x\cos2x}{\sin2x}

    =4\cos2x
    \sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)
    =\sin\frac{\pi}{4}\cos\frac{\theta}{2}-\cos\frac{\pi}{4}\sin\frac{\theta}{2}

     =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)

     =\sqrt{\frac{1}{2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}

     =\sqrt{\frac{1}{2}\left(\cos^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\sin^2\f  rac{\theta}{2}\right)}

     =\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}
    Grandad
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  4. #4
    gp3
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    hi guys, thanks for your help.

    I dont understand how you got this part: =\frac{\sin(3x+x)}{\tfrac12\sin2x}


    and this part: =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right
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  5. #5
    MHF Contributor
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    Hello gp3
    Quote Originally Posted by gp3 View Post
    hi guys, thanks for your help.

    I dont understand how you got this part: =\frac{\sin(3x+x)}{\tfrac12\sin2x}
    \sin (A+B) = \sin A \cos B + \cos A \sin B

    and this part: =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right
    \sin \pi/4 = \cos \pi/4 = 1/\sqrt2

    Grandad
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  6. #6
    gp3
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    erm, i don't understand why is the part in bracket squared?

    =\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}
    and how did you get to 1-Sin?

    sorry for so many questions, i'm rather weak in my maths!
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  7. #7
    MHF Contributor
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    Hello gp3
    Quote Originally Posted by gp3 View Post

    erm, i don't understand why is the part in bracket squared?
     =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)

     =\frac{1}{\sqrt2}\color{red}\sqrt{\left(\color{bla  ck}\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\color{red}\right)^2}

     =\sqrt{\frac{1}{2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}


    =\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}
    and how did you get to 1-Sin?

    sorry for so many questions, i'm rather weak in my maths!
    \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} = 1

    Keep at it - it will come in time!

    Grandad
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