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Thread: Finding a value for tan(7.5°)

  1. #1
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    Finding a value for tan(7.5°)



    I don't really see a way to simplify this any further..
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  2. #2
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    Re: Finding a value for tan(7.5°)

    Quote Originally Posted by pererashenaya7 View Post
    I don't really see a way to simplify this any further..
    That said $4\cdot 7.5^o=30^o$
    $\tan(4x)=?$
    Look at this. Sorry about the size. But you can use the expansion.
    Attached Thumbnails Attached Thumbnails -wolframalpha-simplify_tan_4x___results____2017_03_27_21_19.png  
    Last edited by topsquark; Mar 28th 2017 at 06:38 AM. Reason: But the problem specifies degrees!
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  3. #3
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    Re: Finding a value for tan(7.5°)

    My approach to solving this is as below. I have avoided most of the algebraic manipulations in order to keep this post short and also because I feel you can do them yourself: I have concentrated on the significant steps in my method.

    First, find tan(22½°) using tanA = 2.tan(A/2)/[1 - tan²(A/2)] with A = 45° and remembering that tan(45°) = 1.

    We get: tan(45°) = (2.tan 22½)/(1-tan²22½) = 1
    That produces a quadratic equation in tan(22½), the solutions of which are tan(22½) = -1±√2.

    We also have: tan(A + B) = (tanA + tanB) / (1 - tanA.tanB).
    Let A = 22½; B = 7½ and recall that tan(30°) = 1/√3.
    Substitute those values for A and B in the equation immediately above, plus the calculated value for tan(22½) as just found.
    The resultant algebra gives us tan(7½) = [√3.(√2 - 1) - 1] / [1 - √2 - √3].

    Using a calculator, the RHS evaluates to ± 0.1317; the LHS to ±0.1318. Hence the final result is justified numerically.

    The final problem is transforming [√3.(√2 - 1) - 1] / [1 - √2 - √3] to the format required by the original question. At this point, we are well away from the trig. and algebra: just a question of manipulation of surds.
    For that bit, here, regretfully, I'm stuck!

    Al.
    Last edited by Skywave; Jun 18th 2017 at 05:25 AM.
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  4. #4
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    Re: Finding a value for tan(7.5°)

    Oh, I misinterpreted the question. Ignore my response please.
    Last edited by SlipEternal; Jun 18th 2017 at 08:01 AM.
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