iterated double angle formula — speed of convergence











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The double angle formulas (1) and (2), combined with the best linear approximations at $0$ (3) and (4) can be used to estimate $sin$ or $cos$ by expanding an expression $n$ times and shrinking the angle to $frac{theta}{2^n}$ and then applying the linear approximation.



$$ sin(theta) = 2sin(theta cdot 1/2)cos(theta cdot 1/2) tag{1} $$
$$ cos(theta) = cos^2(theta cdot 1/2) - sin^2(theta cdot 1/2) tag{2} $$



$$ sin(theta) approx theta tag{3} $$
$$ cos(theta) approx 1 tag{4} $$



It seems to converge slowly, maybe adding a bit of accuracy or less per iteration.
How quickly does it converge to the true value of $sin(theta)$ ?



Here's a sample illustrating the relatively slow convergence.



$$
begin{array}[cc]
;n & sin_n(1) \
0 & 1.0 \
1 & 0.9375 \
2 & 0.89233017 \
3 & 0.86740446 \
4 & 0.8545371 \
5 & 0.84802556 \
6 & 0.84475327 \
7 & 0.8431133 \
end{array}
$$










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    The double angle formulas (1) and (2), combined with the best linear approximations at $0$ (3) and (4) can be used to estimate $sin$ or $cos$ by expanding an expression $n$ times and shrinking the angle to $frac{theta}{2^n}$ and then applying the linear approximation.



    $$ sin(theta) = 2sin(theta cdot 1/2)cos(theta cdot 1/2) tag{1} $$
    $$ cos(theta) = cos^2(theta cdot 1/2) - sin^2(theta cdot 1/2) tag{2} $$



    $$ sin(theta) approx theta tag{3} $$
    $$ cos(theta) approx 1 tag{4} $$



    It seems to converge slowly, maybe adding a bit of accuracy or less per iteration.
    How quickly does it converge to the true value of $sin(theta)$ ?



    Here's a sample illustrating the relatively slow convergence.



    $$
    begin{array}[cc]
    ;n & sin_n(1) \
    0 & 1.0 \
    1 & 0.9375 \
    2 & 0.89233017 \
    3 & 0.86740446 \
    4 & 0.8545371 \
    5 & 0.84802556 \
    6 & 0.84475327 \
    7 & 0.8431133 \
    end{array}
    $$










    share|cite|improve this question


























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      The double angle formulas (1) and (2), combined with the best linear approximations at $0$ (3) and (4) can be used to estimate $sin$ or $cos$ by expanding an expression $n$ times and shrinking the angle to $frac{theta}{2^n}$ and then applying the linear approximation.



      $$ sin(theta) = 2sin(theta cdot 1/2)cos(theta cdot 1/2) tag{1} $$
      $$ cos(theta) = cos^2(theta cdot 1/2) - sin^2(theta cdot 1/2) tag{2} $$



      $$ sin(theta) approx theta tag{3} $$
      $$ cos(theta) approx 1 tag{4} $$



      It seems to converge slowly, maybe adding a bit of accuracy or less per iteration.
      How quickly does it converge to the true value of $sin(theta)$ ?



      Here's a sample illustrating the relatively slow convergence.



      $$
      begin{array}[cc]
      ;n & sin_n(1) \
      0 & 1.0 \
      1 & 0.9375 \
      2 & 0.89233017 \
      3 & 0.86740446 \
      4 & 0.8545371 \
      5 & 0.84802556 \
      6 & 0.84475327 \
      7 & 0.8431133 \
      end{array}
      $$










      share|cite|improve this question















      The double angle formulas (1) and (2), combined with the best linear approximations at $0$ (3) and (4) can be used to estimate $sin$ or $cos$ by expanding an expression $n$ times and shrinking the angle to $frac{theta}{2^n}$ and then applying the linear approximation.



      $$ sin(theta) = 2sin(theta cdot 1/2)cos(theta cdot 1/2) tag{1} $$
      $$ cos(theta) = cos^2(theta cdot 1/2) - sin^2(theta cdot 1/2) tag{2} $$



      $$ sin(theta) approx theta tag{3} $$
      $$ cos(theta) approx 1 tag{4} $$



      It seems to converge slowly, maybe adding a bit of accuracy or less per iteration.
      How quickly does it converge to the true value of $sin(theta)$ ?



      Here's a sample illustrating the relatively slow convergence.



      $$
      begin{array}[cc]
      ;n & sin_n(1) \
      0 & 1.0 \
      1 & 0.9375 \
      2 & 0.89233017 \
      3 & 0.86740446 \
      4 & 0.8545371 \
      5 & 0.84802556 \
      6 & 0.84475327 \
      7 & 0.8431133 \
      end{array}
      $$







      trigonometry






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      edited Dec 3 at 6:03

























      asked Dec 3 at 4:02









      Gregory Nisbet

      410311




      410311



























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