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Math Help - Sin Rule

  1. #1
    Member srirahulan's Avatar
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    Sin Rule

    Use only Sin rule and Cos rule for a triangle to proof this>>>>>>> \frac{Cos2A}{a^2}-\frac{Cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}
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  2. #2
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    Re: Sin Rule

    What have you tried so far?
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  3. #3
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    Re: Sin Rule

    Hello, srirahulan!

    We need this identity: . \sin^2\theta \:=\:\frac{1-\cos2\theta}{2}


    \text{Use only Sin rule and Cos rule for a triangle to prove:}

    . . \frac{\cos2A}{a^2}-\frac{\cos2B}{b^2}\:=\:\frac{1}{a^2}-\frac{1}{b^2}

    From the Sine Rule:. . . . \frac{\sin A}{a} \;=\;\frac{\sin B}{b}

    Square both sides:. . . . \frac{\sin^2A}{a^2} \;=\;\frac{\sin^2B}{b^2}


    Apply the identity:. \frac{1-\cos2A}{a^2} \;=\;\frac{1-\cos2B}{b^2}

    . . . . . . . . . . . . . \frac{1}{a^2} - \frac{\cos2A}{a^2} \;=\;\frac{1}{b^2} - \frac{\cos2B}{b^2}

    Therefore: . . \frac{\cos2A}{a^2} - \frac{\cos2B}{b^2} \;=\;\frac{1}{a^2} - \frac{1}{b^2}
    Thanks from topsquark
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