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Math Help - Trigonometric Identity Problem

  1. #1
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    Trigonometric Identity Problem

    How do you show that:
    2sinxcosx=
    (4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
    Thanks a lot for helping!!


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  2. #2
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    Re: Trigonometric Identity Problem

    Quote Originally Posted by JiaGeng View Post
    How do you show that:
    2sinxcosx=
    (4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
    $1+\tan^2(\frac x 2) = \sec^2(\frac x 2)$

    $1-\tan^2(\frac x 2) = \dfrac{2 \tan(\frac x 2)}{\tan(x)}$

    $\sin(2x)=2\sin(x)\cos(x)$

    -------------------------
    $\dfrac{4 \tan(\frac x 2)}{1+\tan^2(\frac x 2)}\dfrac{1-\tan^2(\frac x 2)}{1+\tan^2(\frac x 2)}=$

    $\dfrac{4 \tan(\frac x 2)}{\sec^2(\frac x 2)}\dfrac{ \dfrac{2 \tan(\frac x 2)}{\tan(x)}}{\sec^2(\frac x 2)}=$

    $4\cos(\frac x 2)\sin(\frac x 2)\dfrac{2\sin(\frac x 2)\cos(\frac x 2)}{\tan(x)}=$

    $2\sin(x)\dfrac{\sin(x)}{\tan(x)}=$

    $2\sin(x)\cos(x)$
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  3. #3
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    Re: Trigonometric Identity Problem

    Hello, JiaGeng!

    \text{Prove: }\:2\sin x\cos x\:=\:\frac{4\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \cdot \frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}

    The first fraction is:

    . . \frac{4\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}  }}{\sec^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}  }}{\frac{1}{\cos^2\frac{2}{2}}} \;=\;4\sin\tfrac{x}{2}\cos\tfrac{x}{2} \;=\;2\sin x


    The second fraction is:

    . . \frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\sec^  2\frac{x}{2}} \;=\; \frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac  {1}{\cos^2\frac{x}{2}}} \;=\;\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2} \;=\;\cos x


    Therefore, the RHS becomes: . 2\sin x\cos x.
    Thanks from Ted
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  4. #4
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    Re: Trigonometric Identity Problem

    Thanks romsek and soroban!
    But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
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  5. #5
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    Re: Trigonometric Identity Problem

    Quote Originally Posted by JiaGeng View Post
    Thanks romsek and soroban!
    But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
    Equality works in both directions. If you want, write the equations in reverse order and equality still holds at every step.
    Thanks from romsek
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  6. #6
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    Re: Trigonometric Identity Problem

    Hello, JiaGeng!

    I was wondering if there is a way to do it from LHS to get RHS.
    Would that be even harder?

    If you would rather not run the steps in reverse,
    it can be done ... with a little imagination.

    2\sin x\cos x \;=\;\frac {2(2\sin \frac{x}{2}\cos\frac{x}{2})}{1}\cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{1}

    . . . . . . . . . =\;\frac{4\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\!  \frac{x}{2} + \sin^2\!\frac{x}{2}} \cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2} + \sin^2\!\frac{x}{2}}


    In both fractions, divide numerator and denominator by \cos^2\!\tfrac{x}{2}\!:

    \dfrac{\frac{4\sin\frac{x}{2} \cos\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \cdot \dfrac{\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}  {2}} - \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \;=\; \dfrac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} \cdot \dfrac{1 - \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2}

    . . . . =\;\frac{4\tan\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}} \cdot \frac{1-\tan^2\!\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}}
    Thanks from Ted
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