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    How on earth do I calculate it in exact form??
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    (Original post by XShmalX)
    How on earth do I calculate it in exact form??
    Use the formula for tan(A+B) and remember tan and arctan are inverse functions (subject to appropriate domain/codomain)
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    Expand using the double angle formulae, tan(arctan(3/4)) simplifies to 3/4 and tan(pi/6) is 1/3sqrt3
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    (Original post by ghostwalker)
    Use the formula for tan(A+B) and remember tan and arctan are inverse functions (subject to appropriate domain/codomain)
    Yeah, I tried that but got kinda stuck on what to do with the 1/2... :confused:
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    1/2 is a constant.
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    (Original post by XShmalX)
    Yeah, I tried that but got kinda stuck on what to do with the 1/2... :confused:
    Note that \tan 2x = \dfrac{2\tan x}{1-\tan ^2x}, therefore:

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

    In our case, x=\arctan (\frac{3}{4}) so plugging this in to the above expression we get:

    \dfrac{3}{4} = \dfrac{2\tan (\frac{1}{2}\arctan (\frac{3}{4}))}{1-\tan ^2(\frac{1}{2}\arctan (\frac{3}{4}))}.

    Rearrange this to be a quadratic in terms of \tan (\frac{1}{2}\arctan (\frac{3}{4})) and solve it (remember to get rid of the solution which doesn't make sense). Then follow your method to finish off the problem.
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    (Original post by boromir9111)
    1/2 is a constant.
    I don't think this helps, Boromir?
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    Actually, yeah it doesn't help! sorry about that OP.
 
 
 
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