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Thread: trigonometry proof

  1. #1
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    trigonometry proof

    if tan a=1/2 and tan b=3/8
    (i)find tan(a+b)
    (ii)if tan(a+b+c)=-27 find c given that c is acute.
    i can do part (i) using the trig formula I get (1/2 + 3/8)/[1-(1/2)(3/8)]

    I do not know how to do part (ii) though
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  2. #2
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    Re: trigonometry proof

    Use the same formula for $\tan \left[(a+b) + c\right] = -27$. You will get:

    $-27 = \dfrac{\tan (a+b) + \tan c}{1-\tan (a+b)\tan c}$

    $-27\left[1-\tan (a+b)\tan c\right] = \tan (a+b) + \tan c$

    $27\tan (a+b)\tan c - \tan c = \tan (a+b) + 27$

    $\tan c = \dfrac{\tan (a+b) + 27}{27\tan (a+b) - 1}$

    $c = \arctan \left( \dfrac{\tan (a+b) + 27}{27\tan (a+b) - 1} \right)$

    Edit: I get $c = \dfrac{\pi}{4} = 45^\circ$
    Last edited by SlipEternal; Aug 9th 2017 at 10:23 AM.
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  3. #3
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    Re: trigonometry proof

    ok thanks. so you treat a+b as one term and c as another term.
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