计算三角函数的埃尔密特公式插值法，详细推导过程可见计算三角函数的插值法, 例如。

                                                         (n)                

                    tg`(60°)           tg``(60°)            tg  (60°)  

tg ( 61°)≈tg(60°)+          (61°-60°)+         (61°-60°) +...          (61°-60°)  

                     1!                2!                  n!            

            (N)                                    

          tg  (60°)      n  +1          n  +1           n  +1          

                         0              1               m

       +         (x-x   )        (x-x   )       …  (x-x   )  

            m!      0             1                m

                                                             (n)                

                    tg`(π/3)            tg``(π/3)             tg  (π/3)  

=tg(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...     (180π/61-π/3)  

                     1!                     2!                  n!            

            (3)                                    

          tg  (π/3)          π/3 +1          π/3 +1+1         π/3 +1+1+1                          

       +         (180π/61-π/3)     (180π/61-π/3)       (180π/61-π/3)  

            3!      

                    tg`(π/3)            tg``(π/3)            

≈tg(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...

                     1!                     2!                              

            (1)                                    

          tg  (π/3)          π/3 +1                                  

       +         (180π/61-π/3)    

            1!      

                    tg`(π/3)            tg``(π/3)            

≈tg(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...

                     1!                     2!                              

            2                                    

         sec  (π/3)          π/3 +1                                  

       +         (180π/61-π/3)    

            1!      

                                                           (n)                

                    sin`(60°)           sin``(60°)            sin  (60°)  

sin ( 61°)≈sin(60°)+          (61°-60°)+         (61°-60°) +...          (61°-60°)  

                     1!                2!                  n!            

            (N)                                    

          sin  (60°)      n  +1          n  +1           n  +1          

                         0              1               m

       +         (x-x   )        (x-x   )       …  (x-x   )  

            m!      0             1                m

                                                               (n)                

                    sin`(π/3)            sin``(π/3)             sin  (π/3)  

=sin(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...     (180π/61-π/3)  

                     1!                    2!                  n!            

            (3)                                    

          sin  (π/3)          π/3 +1          π/3 +1+1         π/3 +1+1+1                          

       +         (180π/61-π/3)     (180π/61-π/3)       (180π/61-π/3)  

            3!      

                    sin`(π/3)            sin``(π/3)            

≈sin(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...

                     1!                     2!                              

            (1)                                    

          sin  (π/3)          π/3 +1                                  

       +         (180π/61-π/3)    

            1!      

                    sin`(π/3)            sin``(π/3)            

≈sin(π/3)+        (180π/61-π/3)+       (180π/61-π/3) +...

                     1!                     2!                              

         cos  (π/3)          π/3 +1                                  

       +         (180π/61-π/3)    

            1!      

推导过程见三角函数泰勒级数计算电路中的二项式级数, 计算三角函数的近似公式8

     设set n=10

                    2

        ∞         x

sin x=x*∏ (1-            )

        n=1      2    2

                n   π

                                               2

                                x

                        2                   2                 1-      

                      x                    x         n    -n        2     2

     =x*   2π(1-               )  (  1-              )   e         n    π

                2   2                2    2                      2

n   π               n    π                     x

  1-  

        2

      π  

设set n=10

                     2

        ∞         x

sin x=x*∏ (1-            )

        n=1      2    2

                n   π

                                               2

                                x

                        2                   2                 1-      

                      x                    x         10  -10        2     2

     =10*   2π(1-               )  ( 1-             )   e         10   π

                2   2                2    2                       2

10   π              10  π                       x

  1-  

         2

                                                                     π

                       2

      ∞            4x

ch x=  ∏ (1+                )

      n=1            2    2

                (2n-1)  π

                                                 2

                                  4x

                        2                     2                 1+      

                      4x                    4x         n    -n         2     2

     =     2π(1+               )  (  1+              )   e         (2n-1)    π

                 2   2                  2    2                    2

  (2n-1)  π             (2n-1)   π                   4x

  1+  

        2

      π  

设set n=10

                       2

      ∞            4x

ch x=  ∏ (1+                )

      n=1            2    2

                (2n-1)  π

                                              2

                               4x

                        2                  2                 1+      

                      4x                4x        10  -10           2     2

     =     2π(1+            )  ( 1+              )   e            19    π

               2   2              2    2                      2

  19  π             19   π                      4x

  1+  

       2

    π  

                                 2

π√π=

                                                     1

                                                1-

                    1             1     n   -n         2

            2(1-          )   (1-         ) e          4n

                     2              2               1

   4n            4n             1-

                                    4    

                     2

        ∞         4x

cos x=  ∏ (1-              )

        n=1          2    2

                (2n-1)   π

                                               2

                                4x

                        2                     2                1-      

                      4x                    4x         n    -n         2     2

     =     2π(1-               )  (  1-              )     e       (2n-1)    π

                   2   2                2    2                   2

(2n-1)   π           (2n-1)    π                  4x

  1-  

        2

      π  

设set n=10

                       2

      ∞             4x

cos x=∏ (1-                 )

      n=1            2    2

                (2n-1)   π

                                               2

                                4x

                       2                    2                 1-      

                    4x                   4 x         10    -10        2     2

     =     2π(1-               )  (  1-              )   e         19   π

                2   2                 2    2                       2

19   π               19  π                     4x

  1-  

        2

      π  

                       2

      ∞            x

sh x=x*∏ (1+                )

      n=1            2    2

                   n   π

                                                 2

                                  x

                        2                     2                 1+      

                      x                    x         n    -n         2     2

     = x*   2π(1+               )  (  1+              )   e         (2n-1)    π

                 2   2                2    2                    2

    n   π               n   π                     x

  1+  

        2

      π  

设set n=10

                       2

      ∞            x

sh x=x*∏ (1+                )

      n=1          2    2

                 n    π

                                              2

                                x

                        2                2                   1+      

                      x                x          10   -10           2     2

     =     2π(1+            )  ( 1+              )    e            10    π

               2   2              2    2                      2

  10  π             10   π                       x

  1-  

       2

    π  

计算开方的模拟计算机电路, 推导过程见三角函数泰勒级数计算电路中的二项式级数

              2              

          2z           1     2z     2    1      2z     4     1     2z      6  

   1+ (         )  =1+     (        )   -      (       )   +       (         )-      

       2         2         2       8        2         16        2            

1+  z                 1+ z              1+ z                1+ z  

5        2z      8      n-1  (2n-3)!!   2z     2n-1

      -      (          )  +…+(-1)           (        )  +…       (-1≤x≤1)  

128         2                2n!!         2

1+z                           1+z  

∞    (2n-3)!!     2z      2n-1

=∑        (          )    

                 n=1   2n!!          2

1+z

                    z,如果｜z｜≤1

               ={  1/z,如果if｜z｜≥1

其中：

              2              

          2z        

   1+ (         )    =x  

       2            

1+  z              

         1

                       =  

              2              

          2z          

   1+ (         )      

       2                  

1+  z                

                      1      2z     2    3      2z    4     5      2z      6  

=1-     (       )   +      (       )   -       (         )+      

                 2        2         8       2         16       2            

              1+ z              1+ z                1+ z  

5        2z      8      n-1  (2n-3)!!   2z     2n-1

      -      (          )  +…+(-1)           (        )  +…       (-1≤x≤1)  

128         2                2n!!         2

1+z                           1+z  

∞    (2n-3)!!     2z    2n-1

=∑        (        )    

                 n=1   2n!!          2

1+z

                   z,如果if｜z｜≤1

               ={  1/z,如果if｜z｜≥1

其中：

              2              

          2z        

   1+ (         )    =x  

       2            

1+  z          

         1

                       =  

              2              

          2z          

   1+ (         )      

       2                  

1+  z                

                             2z     2           2z    4            2z      6  

=1-     (       )   +      (       )   -       (         )+      

                           2                2                  2            

              1+ z              1+ z                1+ z  

       2z      8      n      2z     2n-1

      -      (          )  +…+(-1)    (        )  +…       (-1≤x≤1)  

           2                        2

1+z                    1+z  

∞        2z    2n-1

=∑   (        )    

                 n=1           2

1+z

                   z,如果if｜z｜≤1

               ={  1/z,如果if｜z｜≥1

其中：

              2              

          2z        

   1+ (         )    =x  

       2            

1+  z    

推导可以见级数的计算页, 由数学归纳法可得

a        1       1        n-1   1

     =1-      +      -…+(-1)               +…    

b        a       a2           a

其中。a>0,b>0,b-a=1

模拟计算机可以调用这个公式计算除法, 由数学归纳法可得

a        c/2     1         n-1    1

     =1+      +      -…+(-1)            +…    

b        a+b     a          

a

其中。a>0,b>0,b-a=c

由数学归纳法可得

b        1       1         n   1

     =1+      -      -…+(-1)               +…    

a        b       b2           a

其中。a>0,b>0,b-a=1

  模拟计算机可以调用这个公式计算除法, 由数学归纳法可得

b        c/2      c     c        n-1   1

     =1+      -     +      -…+(-1)               +…    

a        a+b     b     b2            b

其中。a>0,b>0,c>0,b-a=c

推导过程可见无穷级数欧拉常数页

       1      1       1  

e=1+       +     +…+      +…

      1!      2!       n!

     ∞

=1+∑

      n

=1+       +C+γ

     n+1      n

其中C=0.57721566490...

用对数函数计算sinx,cosx的公式。推导过程可见无穷级数欧拉常数页

                     2n-1

x

sin x=∑ （-1）              =log(2/π)*x+(2/π)*x*0.001+1+C+γ       (0<x<π/2)

                    (2n-1)!  

    公式(4c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01                        

                     2n-1

x

sin x=∑ （-1）              =log(2/π)*(π-x)+(2/π)*(π-x)*0.001+1+C+γ       (π/2<x<π)

                    (2n-1)!  

    公式(4c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n-1

x

sin x=∑ （-1）              =-log(2/π)*(x-π/2)-(2/π)*(x-π/2)*0.001-1+C+γ      (π<x<3π/2)

                    (2n-1)!  

    公式(4c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n-1

x

sin x=∑ （-1）              =-log(2/π)*(2π-x)-(2/π)*(2π-x)*0.001-1+C+γ      (3π/2<x<2π)

                    (2n-1)!  

    公式(4c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n

x

cos x=∑ （-1）              =-log(2/π)*(π/2-x)+(2/π)*(π/2-x)*0.001+1+C+γ    (0<x<π/2)

                    (2n)!  

    公式(5c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n

x

cos x=∑ （-1）              =-log(1/π)*(x-π/2)-(1/π)*(x-π/2)*0.001-1+C+γ    (π/2<x<π)

                    (2n)!  

    公式(4c)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n

x

cos x=∑ （-1）              =-log(2/π)*(3π/2-x)-(2/π)*(3π/2-x)*0.001-1+C+γ   (π<x<3π/2)

                    (2n)!  

    公式(5a)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

                     2n

x

cos x=∑ （-1）            =log(2/π)*(x-3π/2)+(2/π)*(x-3π/2))*0.01-0.01+1+C+γ (3π/2<x<2π)

                    (2n)!  

    公式(5a)中固定的常数C等于0.01,  这个常数的数值（它是从另外的方法计算出来的）是这样的：   C=0.01    

计算tgx ,ctgx的公式, 推导过程可见级数的乘法页