Using the inverse sec function in Python

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I am creating a program that calculates the optimum angles to fire a projectile from a range of heights and a set initial velocity. Within the final equation I need to utilise, there is an inverse sec function present that is causing some troubles.

I have imported math and attempted to use asec(whatever) however it seems math can not calculate inverse sec functions? I also understand that sec(x) = 1/cos(x) but when I sub 1/cos(x) into the equation instead and algebraically solve for x it becomes a non real result :/.

The code I have is as follows:

print("This program calculates the optimum angles to launch a projectile from a given range of heights and a initial speed.")
x = input("Input file name containing list of heights (m): ")
f = open(x, "r")
for line in f:
    heights = line
    print("the heights you have selected are : ", heights)
f.close()

speed = float(input("Input your initial speed (m/s): "))
print("The initial speed you have selected is : ", speed)

ran0 = speed*speed/9.8 
print(ran0)

f = open(x, "r")
for line in f:
    heights = (line)
    import math
    angle = (math.asec(1+(ran0/float(heights))))/2
    print(angle)

f.close()

So my main question is, is there any way to find the inverse sec of anything in python without installing and importing something else? I realise this may be more of a math based problem than a coding problem however any help is appreciated :).

math python
4 Answers

Let's say we're looking for real number x whose arcsecant is angle θ. Then we have:

θ = arcsec(x)
sec(θ) = x
1 / cos(θ) = x
cos(θ) = 1 / x
θ = arccos(1/x)

So with this reasoning, you can write your arcsecant function as:

from math import acos

def asec(x):
    return acos(1/x)
user_93321870
user_128067620

"I also understand that sec(x) = 1/cos(x) but when I sub 1/cos(x) ..." Do you have to use sec or asec ?

Because sec(x)= 1/cos(x) and asec(x) = acos(1/x). Be careful the notation ^-1 is ambiguous, cos^-1(x) = acos(x) is different of [cos(x)]^-1.

angle = (math.asec(1+(ran0/float(heights))))/2

asec is not defined from -1 to 1 If you have a height lower than zero, and so the result of (ran0/float(heights)) is between -2 and 0, your angle will be non real.

I don't really know if this is what you asked for, but I hope it helps.

user_95633870

If math is OK for you to import, then you can use:

import math


def asec(x):
    if x == 0:
        return 1j * math.inf
    else:
        return math.acos(1 / x)

For some other ways of of re-writing asec(x), feast your eyes on the relevant Wikipedia article.

Alternatively, you could use Taylor series expansions, which always come in polynomial form, so, although that is only an approximation in a neighborhood of a given point, it would not require math.

For asec(x), its Taylor expansion in a neighborhood of ±∞ (also known as Laurent series) is given by (without using math):

def asec_taylor(x, pi=3.14159265):
    if x == 0:
        return 1j * float('inf')
    else:
        return pi / 2 - 1 / x - 1 / (6 * x ** 3) - 3 / (40 * x ** 5)

You can quickly check that the farther you are from 0, the better the approximation holds:

for x in range(-10, 10):
    print(x, asec(x), asec_taylor(x))

-10 1.6709637479564565 1.670963741666667
-9 1.6821373411358604 1.6821373299281108
-8 1.696124157962962 1.6961241346516926
-7 1.714143895700262 1.7141438389326868
-6 1.7382444060145859 1.7382442416666668
-5 1.7721542475852274 1.7721536583333335
-4 1.8234765819369754 1.823473733854167
-3 1.9106332362490186 1.910611139814815
-2 2.0943951023931957 2.0939734083333335
-1 3.141592653589793 2.8124629916666666
0 (nan+infj) (nan+infj)
1 0 0.32912965833333346
2 1.0471975511965979 1.0476192416666668
3 1.2309594173407747 1.2309815101851853
4 1.318116071652818 1.3181189161458333
5 1.369438406004566 1.3694389916666667
6 1.4033482475752073 1.4033484083333334
7 1.4274487578895312 1.4274488110673134
8 1.4454684956268313 1.4454685153483076
9 1.4594553124539327 1.4594553200718894
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