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How do you do a sign function?

  1. Oct 5, 2005 #1

    box

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    I have always used a calculator to find the sin when the angle is given. But how would you figure this out if you didn't have a calculator
     
  2. jcsd
  3. Oct 5, 2005 #2

    arildno

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    Look up in a sine table.
     
  4. Oct 5, 2005 #3

    krab

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    [tex]\sin x=x-x^3/6+x^5/120-...[/tex]
    One can use the symmetries of the function to avoid large values of x (where convergence is slow) and only ever have to calculate with x as large as pi/2. For example, sin(2)=sin(pi-2)=sin(1.14159..). One can even do better by using the expansion of cosine:
    [tex]\cos x=1-x^2/2+x^4/24-...[/tex]
    Then using sin^2+cos^2=1, you only need the series up to x=pi/4. The series converges very quickly for such small x.
     
  5. Oct 5, 2005 #4
    ...and to a limited extent you can calculate them by hand using the angles you already know (30, 45, 60, 90, etc.) with identities such as the double and half-angle formulas.
     
  6. Oct 6, 2005 #5

    HallsofIvy

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    You can use the Taylor's series for sin(x)= x- x2/2+ x4/4!- x6/6!+ ... where the general term is (-1)nx2n/(2n)! and a few terms should get you pretty good accuracy.

    My understanding is that calculators and computers actually use the "CORDIC" method:
    http://www.dspguru.com/info/faqs/cordic.htm
     
  7. Oct 6, 2005 #6

    VietDao29

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    There's a slight error there... You just mixed up the Taylor expansion for sin(x), and cos(x) :smile:.
    [tex]\sin x = x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - \frac{x ^ 7}{7!} + ... = \sum_{n = 0} ^ {\infty} \frac{(-1) ^ n}{(2n + 1)!} x ^ {2n + 1}[/tex]
    [tex]\cos x = 1 - \frac{x ^ 2}{2!} + \frac{x ^ 4}{4!} - \frac{x ^ 6}{6!} + ... = \sum_{n = 0} ^ {\infty} \frac{(-1) ^ n}{(2n)!} x ^ {2n}[/tex]
    Viet Dao,
     
  8. Oct 7, 2005 #7

    HallsofIvy

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    Oops!

    No wonder I keep getting my trig problems wrong!
     
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