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How to get an exact answer instead of a decimal? - Trig+complex no. question in post4

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    Hello,

    There's a question from a past papers which wants the answer to be given as an exact value. Answer in mark scheme is 2 + \sqrt{3} and it can be obtained by calculating 2\sqrt{5}[\cos (\pi - \tan^{-1} (0.5) - \frac{2}{3}\pi)]. My calculator gives me the answer in form of 3.732050808. I have no idea how to get it into an exact form...

    ...any help would be appreciated!

    Original question is part (iv) in post#4!
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    That tan^{-1}(0.5) is what's making this hard. Hmm an A level exam asks for this?

    What was the original question?
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    (Original post by Zishi)
    Hello,

    There's a question from a past papers which wants the answer to be given as an exact value. Answer in mark scheme is 2 + \sqrt{3} and it can be obtained by calculating 2\sqrt{5}[\cos (\pi - \tan^{-1} (0.5) - \frac{2}{3}\pi)]. My calculator gives me the answer in form of 3.732050808. I have no idea how to get it into an exact form...

    ...any help would be appreciated!
    Any chance we can see the question? It's possible there's another route ...
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    (Original post by Jam')
    That tan^{-1}(0.5) is what's making this hard. Hmm an A level exam asks for this?

    What was the original question?
    (Original post by ian.slater)
    Any chance we can see the question? It's possible there's another route ...
    Yes, of course there's another route - I did it using an argand diagram(It's a question from Complex Numbers), whereas the mark scheme used 'analogy' from part (ii) to solve part(iv):


    Mark Scheme:


    I now do understand the way mark scheme used, but while solving the question, the only way that popped into my mind was to do it using an argand diagram. But as stated in post#1, that way doesn't give an exact answer....
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    Yes, in general rotating a point by converting to polar coordinates, adding the rotation angle and converting back won't give something that's has an obvious exact form.

    If you desperately wanted to go from your equation involving tan, you could rewrite as cos(t + pi /3), where t=arctan(0.5).

    Tan (pi/3) = sqrt(3), so tan(t+pi/3) = (\sqrt(3)+1/2)/(1-\sqrt{3}/2). Then use cos^2 A = 1/(1+tan^2 A). Wouldn't be a lot of fun, though.
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    (Original post by DFranklin)
    Yes, in general rotating a point by converting to polar coordinates, adding the rotation angle and converting back won't give something that's has an obvious exact form.

    If you desperately wanted to go from your equation involving tan, you could rewrite as cos(t + pi /3), where t=arctan(0.5).

    Tan (pi/3) = sqrt(3), so tan(t+pi/3) = (\sqrt(3)+1/2)/(1-\sqrt{3}/2). Then use cos^2 A = 1/(1+tan^2 A). Wouldn't be a lot of fun, though.
    WOW, I never thought I could use such an easy method. Thanks a lot for it!

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