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Math Help - Trig identity proof

  1. #1
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    Trig identity proof

    prove that  tan50^{o} -tan40^{o} = 2tan10^{o}
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  2. #2
    MHF Contributor red_dog's Avatar
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    Use these formulas:

    \tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}

    and \cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]
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  3. #3
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    Quote Originally Posted by red_dog View Post
    Use these formulas:

    \tan a-\tan b=\frac{\sin(a-b)}{\cos a\cos b}

    and \cos a\cos b=\frac{1}{2}[\cos(a+b)+\cos(a-b)]
    thanks,

    but I have never seen these formula's or been taught them. where did you get them from?

    surely If I needed to use these formula's It would be in the book, that the question is from?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by Tweety View Post
    thanks,

    but I have never seen these formula's or been taught them. where did you get them from?

    surely If I needed to use these formula's It would be in the book, that the question is from?

    Prove that \tan 50 - \tan 40 = 2\tan 10
    Hi Tweety,

    I don't know what formulas you have been taught, but here's another proof using the Tan sum formula:

    \tan(u+v)=\frac{\tan u+\tan v}{1-\tan u \tan v}.



    \tan (50) = \tan (40+10)= \frac{\tan 40 + \tan 10}{1- \tan 40 \tan 10}

    \tan 50 (1- \tan 40 \tan 10) = \tan 40 + \tan 10<br />

    Now, distribute:

    \tan 50 - {\color{red}\tan 50 \tan 40} \tan 10 = \tan 40 + \tan 10

    You know that {\color{red}\tan 50 \tan 40 = 1}

    \tan 50 - \tan 10= \tan 40 + \tan 10

    \tan 50 = \tan 40 + 2 \tan 10

    \tan 50 - \tan 40 = 2 \tan 10
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