Simplify trigonometric and inverse trigonometric functions using linear interpolation

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

sin

In the following example, we first define the interpolation range (from 0 to π/2), and then calculate the interpolation step size according to the accuracy requirement. Then, we find the interpolation point closest to the input value and use the linear interpolation formula to approximate the value of the sin function.

Note that the higher the accuracy requirement (i.e. the number of interpolation points), the closer the approximation will be to the exact value of the sin function at the given input values. However, the approximation may have limited accuracy due to the use of linear interpolation. If higher precision approximation results are required, higher order interpolation methods such as quadratic or cubic interpolation may be considered.

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_sin(x, precision):
    """
    Approximate the sin function using linear interpolation
    """
    # define interpolation range
    x0 = 0.0
    y0 = math. sin(x0)
    x1 = math.pi / 2
    y1 = math. sin(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.sin(interpolation_point - step),
                                         interpolation_point, math. sin(interpolation_point))
    return approximation

# Example usage
x = 1.0 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_sin(x, precision)
print(approximation)

cos

In the above example, we define the interpolation range and precision requirements in a similar way, then find the interpolation point closest to the input value, and use the linear interpolation formula to approximate the value of the cos function.

Likewise, the accuracy of the approximation depends on the accuracy requirement (number of interpolated points). If higher precision approximation results are required, higher order interpolation methods such as quadratic or cubic interpolation may be considered.

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_cos(x, precision):
    """
    Approximating the cos function using linear interpolation
    """
    # define interpolation range
    x0 = 0.0
    y0 = math.cos(x0)
    x1 = math.pi / 2
    y1 = math.cos(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.cos(interpolation_point - step),
                                         interpolation_point, math.cos(interpolation_point))
    return approximation

# Example usage
x = 1.0 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_cos(x, precision)
print(approximation)

tan

For the linear interpolation approximation of the tan function, it should be noted that the tan function will diverge at some points (such as multiples of π/2), so the interpolation range needs to exclude these points. Here is the code for an example linear interpolation approximation to the tan function:

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_tan(x, precision):
    """
    Approximate the tan function using linear interpolation
    """
    # Define the interpolation range (excluding points where the tan function diverges)
    x0 = -math.pi / 2 + 0.01
    y0 = math.tan(x0)
    x1 = math.pi / 2 - 0.01
    y1 = math.tan(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.tan(interpolation_point - step),
                                         interpolation_point, math.tan(interpolation_point))
    return approximation

# Example usage
x = 1.0 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_tan(x, precision)
print(approximation)

arcsin

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_arcsin(x, precision):
    """
    Approximate arcsin function using linear interpolation
    """
    # define interpolation range
    x0 = -1.0
    y0 = math.asin(x0)
    x1 = 1.0
    y1 = math.asin(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.asin(interpolation_point - step),
                                         interpolation_point, math.asin(interpolation_point))
    return approximation

# Example usage
x = 0.5 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_arcsin(x, precision)
print(approximation)

In the above example, we defined the interpolation range (from -1 to 1), and calculated the interpolation step size. Then, find the interpolation point closest to the input value and use the linear interpolation formula to approximate the value of the arcsin function.

Note that due to the nature of the arcsin function, the approximation may be less accurate at some points, so an appropriate interpolation range and accuracy requirement needs to be chosen. If higher accuracy approximation results are required, other higher order interpolation methods or other approximation techniques may be considered.

arccos

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_arccos(x, precision):
    """
    Approximating an arccos function using linear interpolation
    """
    # define interpolation range
    x0 = -1.0
    y0 = math. acos(x0)
    x1 = 1.0
    y1 = math. acos(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.acos(interpolation_point - step),
                                         interpolation_point, math.acos(interpolation_point))
    return approximation

# Example usage
x = 0.5 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_arccos(x, precision)
print(approximation)

In the above example, we defined the interpolation range (from -1 to 1), and calculated the interpolation step size. Then, find the interpolation point closest to the input value and use the linear interpolation formula to approximate the value of the arccos function.

Similar to the approximation of other trigonometric functions, due to the nature of the arccos function, the approximation may be less accurate at some points. Therefore, for higher precision requirements, other higher order interpolation methods or other approximation techniques may be considered.

arctan

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_arctan(x, precision):
    """
    Approximate arctan function using linear interpolation
    """
    # define interpolation range
    x0 = -math.pi/2
    y0 = math.atan(x0)
    x1 = math.pi / 2
    y1 = math.atan(x1)

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < x:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(x, interpolation_point - step, math.atan(interpolation_point - step),
                                         interpolation_point, math.atan(interpolation_point))
    return approximation

# Example usage
x = 1.0 # input value
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_arctan(x, precision)
print(approximation)

arctan2

import math

def linear_interpolation(x, x0, y0, x1, y1):
    """
    Linear interpolation function
    """
    return y0 + (y1 - y0) * (x - x0) / (x1 - x0)

def approximate_arctan2(y, x, precision):
    """
    Approximating the arctan2 function using linear interpolation
    """
    # calculate arctan value
    arctan_yx = math. atan2(y, x)

    # Define the interpolation range (usually [-pi, pi])
    x0 = -math.pi
    y0 = math. atan2(math. sin(x0), math. cos(x0))
    x1 = math.pi
    y1 = math. atan2(math. sin(x1), math. cos(x1))

    # Calculate interpolation step size
    step = (x1 - x0) / precision

    # find the interpolation point
    interpolation_point = x0
    while interpolation_point < arctan_yx:
        interpolation_point += step

    # do linear interpolation
    approximation = linear_interpolation(arctan_yx, interpolation_point-step,
                                         math.atan2(math.sin(interpolation_point - step), math.cos(interpolation_point - step)),
                                         interpolation_point,
                                         math.atan2(math.sin(interpolation_point), math.cos(interpolation_point)))
    return approximation

# Example usage
y = 1.0 # y coordinate
x = 1.0 # x coordinate
precision = 1000 # precision requirement, that is, the number of interpolation points
approximation = approximate_arctan2(y, x, precision)
print(approximation)

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