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Math Help - Prove:

  1. #1
    Member great_math's Avatar
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    Prove:

    If \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}


    Prove that \sin2a+\sin2b+\sin2c=0
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  2. #2
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    Quote Originally Posted by great_math View Post
    If \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}


    Prove that \sin2a+\sin2b+\sin2c=0
    Hi

    Let a=b=c=\frac{\pi}{4}

    Then \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan\left(\frac{\pi}{4}\right)}{\tan\l  eft(\frac{\pi}{4}\right)}=1

    And \frac{\tan c}{\tan b} = 1

    Therefore \frac{\tan(a+b-c)}{\tan(a-b+c)}=\frac{\tan c}{\tan b}

    But \sin2a+\sin2b+\sin2c=1+1+1=3
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Hey that's not what you were supposed to prove!
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    Hey that's not what you were supposed to prove!
    When something seems very difficult to prove I prefer spending some time trying to find a counter example showing that it is false !
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