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Math Help - Proving a Trigonometric Identity

  1. #1
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    Proving a Trigonometric Identity

    Hi everyone:

    Is there any way to prove the following identity:

    tan(A/2)=(sinA)/(1+cosA)

    without drawing a diagram? I know that it is a basic trig identity, but I don't know how to prove it by manipulating hte formulas.

    Thanks!
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  2. #2
    Super Member bigwave's Avatar
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    Cool use sin(x/2) / cos(x/2)

    \tan{\frac{A}{2}}=\frac{\sin{A}}{1+\cos{A}}

    this derived from by using

    \tan{\frac{A}{2}}=\frac{\sin\frac{A}{2}}{\cos\frac  {A}{2}}

    <br />
\frac<br />
{\pm\sqrt{\frac{1-\cos{A}}{2}}}<br />
{\pm\sqrt{\frac{1+\cos{A}}{2}}}<br />

    =\pm\sqrt\frac{\left(1-\cos{A}\right)\times\left(1+cos{A}\right)}<br />
{\left(1+\cos{A}\right)\times\left(1+\cos{A}\right  )}<br />

    =\pm\sqrt\frac{1-\cos^2{A}}<br />
{\left(1+\cos{A}\right)^2}<br />

    =\pm\vert\frac{\sin{A}}{1-\cos{A}}\vert

    1-\cos{A} in never negative, so the sign of the fractional expression depends only on the sign of \sin{A}
    Last edited by bigwave; December 21st 2009 at 10:25 AM. Reason: finish post
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  3. #3
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    Hello, Kelvin!
    We need these two identities: . \begin{array}{ccc}\sin\frac{\theta}{2} &=& \sqrt{\dfrac{1-\cos\theta}{2}} \\ \\[-4mm]\cos\frac{\theta}{2} &=& \sqrt{\dfrac{1+\cos\theta}{2}} \end{array}


    Prove: . \tan\frac{A}{2} \:=\:\frac{\sin A}{1+\cos A}

    \tan\frac{A}{2} \;=\;\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \;=\;\frac{\sqrt{\dfrac{1-\cos A}{2}}} {\sqrt{\dfrac{1 + \cos A}{2}}} \;=\; \sqrt{\frac{1-\cos A}{1+\cos A}}


    Multiply by \frac{1+\cos A}{1 + \cos A}

    . . \sqrt{\frac{1-\cos A}{1+\cos A}\cdot\frac{1+\cos A}{1 + \cos A}} \;=\;\sqrt{\frac{1-\cos^2\!A}{(1+\cos A)^2}} \;=\;\sqrt{\frac{\sin^2\!A}{(1+\cos A)^2}} \;=\; \frac{\sin A}{1 + \cos A}

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  4. #4
    Senior Member I-Think's Avatar
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     tan(\frac{A}{2})=\frac{sinA}{1+cosA}

    Manipulating R.H.S.
    \frac{2sin(\frac{A}{2}){cos(\frac{A}{2})}}{1+cos^2  (\frac{A}{2})-sin^2(\frac{A}{2})}

    Note that
    1=sin^2(\frac{A}{2})+cos^2(\frac{A}{2})

    Introduce into the equation and you should solve your problem

    Edit
    In the time I took to write this I was beaten by 2 other forum users
    I really need to take typing classes.
    On the other hand, at least I provided a different method
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  5. #5
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    Thank you so much!
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  6. #6
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    Okay, Similar Question.

    Prove that:

    tan(A/2)=(1+sinA-cosA)/(1+sinA+cosA)



    I tried replacing tan(A/2) with sinA/(1+cosA), but I could not find a way to add the extra components to both the numerator and denominator. Gosh this is frustrating.
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  7. #7
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    Helolo, Kelvin!

    This one is tricky . . .


    Prove: . \tan\frac{A}{2} \;=\;\frac{1+\sin A-\cos A}{1+\sin A+\cos A}
    Start with the right side . . .


    Multiply by \frac{1+\sin A - \cos A}{1 + \sin A - \cos A}

    . . \frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}\cdot{\color{blue}\frac{1 + \sin A - \cos A}{1 + \sin A - \cos A}} \;=\;\frac{(1+\sin A - \cos A)^2}{(1+\sin A)^2 - \cos^2\!A}

    . . =\; \frac{1 + 2\sin A - 2\cos A + \sin^2\!A - 2\sin A\cos A + \cos^2\!A}{1 + 2\sin A + \sin^2\!A - \cos^2\!A}

    . . =\; \frac{1 + \overbrace{\sin^2\!A + \cos^2\!A}^{\text{This is 1}} + 2\sin A - 2\cos A - 2\sin A\cos A}{2\sin A + \sin^2\!A + \underbrace{1 - \cos^2\!A}_{\text{This is }\sin^2\!A}}

    . . =\; \frac{2 + 2\sin A - 2\cos A - 2\sin A\cos A}{2\sin A + 2\sin^2\!A} \;=\;\frac{2(1 + \sin A - \cos A - \sin A\cos A}{2\sin A(1 + \sin A)}<br />

    . . =\; \frac{1 + \sin A - \cos A - \sin A\cos A}{\sin A(1 + \sin A)} \;=\;\frac{(1 + \sin A) - \cos A(1 + \sin A)}{\sin A(1 + \sin A)}

    . . =\; \frac{(1+\sin A)(1 - \cos A)}{\sin A(1 + \sin A)} \;=\;\frac{1-\cos A}{\sin A}


    Multiply by \frac{1+\cos A}{1 + \cos A}\!:\quad \frac{1-\cos A}{\sin A}\cdot{\color{blue}\frac{1+\cos A}{1+\cos A}} \;=\; \frac{1-\cos^2\!A}{\sin A(1 + \cos A)}

    . . . . . . . . . . . . . =\;\frac{\sin^2\!A}{\sin A(1 + \cos A)} \;=\;\frac{\sin A}{1 + \cos A}


    Finally: . \frac{\sin A}{1 + \cos A} \;=\;\tan\frac{A}{2} \quad \hdots\;There!

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  8. #8
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    be careful with what angle you're working on, those identities are not true for all angles.
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