DuckieDai Inverse Trig Algebra Python Program

trigonometry · MIT License

Overview

Write an expression such as cos(sin^-1(12x)) or tan(sec^-1(x/3)) as an algebraic expression in x, with the reference triangle and the answer rationalized.

Owner-tested on a TI calculator
Version 1.0.0
2172 bytes
No imports

Download INVALG.py Read the source How to install

Before you run it

TI-84 Plus CE Python · No calculator OS or Python App version has been verified.
TI Connect CE .py transfer to the Python App; the owner's transfer details were not recorded
The inside of the inverse is a positive multiple of x over a whole number or a square root, such as 12x, x/3, 2x/5, or x/sqrt5.
The triangle is drawn for positive x, so the sides are the usual textbook ones.
The inside must contain x once, with positive whole numbers: 12x, x/3, 2x/5, or x/sqrt5.

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Learn the math

How to write cos(sin^-1(12x)) and similar expressions in terms of x

The idea

An expression such as cos(sin^-1(12x)) asks for the cosine of the angle whose sine is 12x. You do not need the angle. Treat 12x as a ratio, opposite over hypotenuse, draw a right triangle with those sides, and find the third side with the Pythagorean theorem.

Then read the outside function off the triangle as a ratio of two sides. When the answer has a square root in the denominator, multiply the top and bottom by that square root to rationalize it.

The formula

sin^-1(a/b): opp = a, hyp = b, adj = sqrt(b^2 - a^2)

tan^-1(a/b): opp = a, adj = b, hyp = sqrt(a^2 + b^2)

x/sqrt(x^2+9) = x*sqrt(x^2+9)/(x^2+9)

reference triangle: the right triangle built from the inside ratio
rationalize: rewrite a fraction so no square root is left in the denominator
x/sqrt5: how to type x over the square root of 5

Using the program

Choose the outside function, 1 sin to 6 cot, then the inside inverse, 1 sin^-1 to 6 cot^-1.
Type the inside as a multiple of x over a whole number or a square root, such as 12x, x/3, 2x/5, or x/sqrt5. Fractions are reduced first, so 4x/6 becomes 2x/3.
The program shows the inside, the triangle's opp, adj, and hyp, and the ratio the outside function uses.
The ANSWER comes last. When a square root was in the denominator, the program first shows the form before rationalizing.

Worked example

Write cos(sin^-1(12x)) in terms of x. Choose 2 for cos, 1 for sin^-1, and type 12x.

opp = 12x, hyp = 1, adj = sqrt(1 - 144x^2), cos = adj/hyp

The program shows Triangle: opp 12x, adj sqrt(1-144x^2), hyp 1, then cos = adj/hyp and ANSWER: sqrt(1-144x^2).

For sin(tan^-1(x/3)), choose 1, 3, and type x/3. The triangle has opp x, adj 3, and hyp sqrt(x^2+9). The program shows Before rationalizing: x/sqrt(x^2+9), then the answer xsqrt(x^2+9)/(x^2+9).

For sec(cos^-1(2x/5)), choose 5, 2, and type 2x/5. The adjacent side is 2x and the hypotenuse 5, so the program shows sec = hyp/adj and the answer 5/(2x).

Tips and common mistakes

Decide which two sides the inside gives you: sin^-1 and csc^-1 use the opposite and hypotenuse, cos^-1 and sec^-1 the adjacent and hypotenuse, tan^-1 and cot^-1 the two legs.
The hypotenuse is always the longest side, so for sin^-1 and cos^-1 the missing side has a minus sign under the root: b^2 - a^2.
Read the program's xsqrt(x^2+9) as x times sqrt(x^2+9).
These answers assume x is positive, as textbooks usually do; for negative x, check the quadrant the angle lands in.

Where this comes up

Rewriting a trig function of an inverse trig function in terms of x is part of the inverse trig unit of precalculus, and calculus uses it with trig substitution in integrals.

Exact reviewed bytes

Source code

Download .py

View source below, copy it when JavaScript is available, or download the exact .py bytes.

def clr():print("\n"*11)
def pg():input("[enter]");clr()
def pick(p,v):
 while 1:
  c=input(p).strip()
  if len(c)==1 and c in v:return c
clr();print("INVERSE TRIG ALGEBRA\n like cos(sin^-1(12x))\n      __\n  ___(o )>  quack!\n  \\ <_. )\n   `---'\nDuckieDai (TM)\nhiddenwordscanner.com");pg()
F=("sin","cos","tan","csc","sec","cot")
# inverse -> (opp,adj,hyp,radical type)
T=(("K","R","B",1),("R","K","B",1),("K","B","R",2),("B","R","K",3),("R","B","K",3),("B","K","R",2))
while 1:
 o=pick("OUTSIDE:\n1 sin  2 cos  3 tan\n4 csc  5 sec  6 cot\n0 Exit\n> ","0123456")
 if o=="0":break
 i=int(pick("INSIDE inverse:\n1 sin^-1  2 cos^-1\n3 tan^-1  4 csc^-1\n5 sec^-1  6 cot^-1\n> ","123456"))-1
 s=input("Inside (12x, x/3,\n2x/5, x/sqrt5)? ").replace(" ","").replace("(","").replace(")","").lower()
 try:
  n,d=(s.split("/")+["1"])[:2];n=n.replace("x","",1)
  if "x" not in s or "x" in n+d:raise ValueError
  a=int(n) if n else 1;q=0
  if d[:4]=="sqrt":q=int(d[4:]);b=1
  else:b=int(d)
  if a<1 or b<1 or q<0:raise ValueError
 except ValueError:print("Type like 12x  x/3\nor x/sqrt5");pg();continue
 r=int(q**.5+.5)
 if r*r==q:b=r or b;q=0
 g,h=a,b
 while h:g,h=h,g%h
 if q:g=1
 a//=g;b//=g;K=("" if a==1 else str(a))+"x";Q="sqrt("+str(q)+")"
 op,ad,hy,t=T[i];
 def X(m):A=a*a*m;A="" if A==1 else str(A);B=str((q or b*b)*m);return (B+"-"+A+"x^2",A+"x^2+"+B,A+"x^2-"+B)[t-1]
 E=X(1);R="sqrt("+E+")"
 def S(z):return K if z=="K" else ((Q if q else str(b)) if z=="B" else R)
 clr();print(F[i]+"^-1("+K+("/"+Q if q else "/"+str(b) if b>1 else "")+")\nTriangle: opp "+S(op)+"\n adj "+S(ad)+"\n hyp "+S(hy))
 o=int(o)-1;nu,de=((op,hy),(ad,hy),(op,ad),(hy,op),(hy,ad),(ad,op))[o];N=S(nu)
 if de=="B" and q:r=("sqrt("+X(q)+")" if N==R else N+Q)+"/"+str(q);print("Before rationalizing:\n "+N+"/"+Q)
 elif de=="B":r=N if b==1 else N+"/"+str(b)
 elif de=="K":r=N+"/"+(K if a==1 else "("+K+")")
 else:
  top=R if N=="1" else ("sqrt("+X(q)+")" if q and N==Q else N+R)
  r=top+"/("+E+")";print("Before rationalizing:\n "+N+"/"+R)
 print(F[o]+" = "+("opp/hyp","adj/hyp","opp/adj","hyp/opp","hyp/adj","adj/opp")[o]+"\nANSWER:\n "+r);pg()
print("Quack! DuckieDai (TM)\nhiddenwordscanner.com")

def clr():print("\n"*11)
def pg():input("[enter]");clr()
def pick(p,v):
while 1:
c=input(p).strip()
if len(c)==1 and c in v:return c
clr();print("INVERSE TRIG ALGEBRA\n like cos(sin^-1(12x))\n __\n ___(o )> quack!\n \\ <_. )\n `---'\nDuckieDai (TM)\nhiddenwordscanner.com");pg()
F=("sin","cos","tan","csc","sec","cot")
# inverse -> (opp,adj,hyp,radical type)
T=(("K","R","B",1),("R","K","B",1),("K","B","R",2),("B","R","K",3),("R","B","K",3),("B","K","R",2))
while 1:
o=pick("OUTSIDE:\n1 sin 2 cos 3 tan\n4 csc 5 sec 6 cot\n0 Exit\n> ","0123456")
if o=="0":break
i=int(pick("INSIDE inverse:\n1 sin^-1 2 cos^-1\n3 tan^-1 4 csc^-1\n5 sec^-1 6 cot^-1\n> ","123456"))-1
s=input("Inside (12x, x/3,\n2x/5, x/sqrt5)? ").replace(" ","").replace("(","").replace(")","").lower()
try:
n,d=(s.split("/")+["1"])[:2];n=n.replace("x","",1)
if "x" not in s or "x" in n+d:raise ValueError
a=int(n) if n else 1;q=0
if d[:4]=="sqrt":q=int(d[4:]);b=1
else:b=int(d)
if a<1 or b<1 or q<0:raise ValueError
except ValueError:print("Type like 12x x/3\nor x/sqrt5");pg();continue
r=int(q**.5+.5)
if r*r==q:b=r or b;q=0
g,h=a,b
while h:g,h=h,g%h
if q:g=1
a//=g;b//=g;K=("" if a==1 else str(a))+"x";Q="sqrt("+str(q)+")"
op,ad,hy,t=T[i];
def X(m):A=a*a*m;A="" if A==1 else str(A);B=str((q or b*b)*m);return (B+"-"+A+"x^2",A+"x^2+"+B,A+"x^2-"+B)[t-1]
E=X(1);R="sqrt("+E+")"
def S(z):return K if z=="K" else ((Q if q else str(b)) if z=="B" else R)
clr();print(F[i]+"^-1("+K+("/"+Q if q else "/"+str(b) if b>1 else "")+")\nTriangle: opp "+S(op)+"\n adj "+S(ad)+"\n hyp "+S(hy))
o=int(o)-1;nu,de=((op,hy),(ad,hy),(op,ad),(hy,op),(hy,ad),(ad,op))[o];N=S(nu)
if de=="B" and q:r=("sqrt("+X(q)+")" if N==R else N+Q)+"/"+str(q);print("Before rationalizing:\n "+N+"/"+Q)
elif de=="B":r=N if b==1 else N+"/"+str(b)
elif de=="K":r=N+"/"+(K if a==1 else "("+K+")")
else:
top=R if N=="1" else ("sqrt("+X(q)+")" if q and N==Q else N+R)
r=top+"/("+E+")";print("Before rationalizing:\n "+N+"/"+R)
print(F[o]+" = "+("opp/hyp","adj/hyp","opp/adj","hyp/opp","hyp/adj","adj/opp")[o]+"\nANSWER:\n "+r);pg()
print("Quack! DuckieDai (TM)\nhiddenwordscanner.com")

Filename: INVALG.py
Size: 2172 bytes
SHA-256: 21b3eafc0c54514c64e16e72f79baff7a909da40a3ebc42b4a4394723ae706ee

Usage

Inputs

The outside function: 1 sin, 2 cos, 3 tan, 4 csc, 5 sec, 6 cot, or 0 to exit
The inside inverse: 1 sin^-1 to 6 cot^-1
The inside, such as 12x, x/3, 2x/5, or x/sqrt5

Outputs

The reference triangle's opposite, adjacent, and hypotenuse in terms of x
The ratio the outside function uses, such as cos = adj/hyp
The answer, rationalized, with the form before rationalizing when there was a square root in the denominator

Units: Algebraic expressions in x; no units.

Assumptions and limitations

Assumptions

The inside of the inverse is a positive multiple of x over a whole number or a square root, such as 12x, x/3, 2x/5, or x/sqrt5.
The triangle is drawn for positive x, so the sides are the usual textbook ones.

Constraints

The inside must contain x once, with positive whole numbers: 12x, x/3, 2x/5, or x/sqrt5.
Fractions are reduced, so 4x/6 becomes 2x/3, and x/sqrt9 becomes x/3.

Known failures

Negative coefficients, decimals, and expressions such as x+1 are not accepted.
A square root typed with more than 300 digits stops the program with an error.

Compatibility and review

Review status: Owner-tested on a TI calculator
Tested on: Owner reports personally running this program on a TI-84 Plus CE Python calculator; the test date, OS version, and Python App version were not recorded
What happened: Owner reported pass on the calculator. The published file is the owner's tested source.
Dependencies: No imports
Suggested calculator name: INVALG — you can give it another valid, unique name when you transfer it
Desktop test cases: Not available for this interactive-only source

This exact source auto-launches an interactive calculator session, so compatibility evidence comes from the recorded physical-device review rather than an importable desktop fixture.

Transfer and launch

Download the .py file, send it to your calculator with TI Connect CE as INVALG or another name, then open the Python App and try the worked example on this page first. This release has no verified calculator OS or Python App version; treat transfer and execution as unverified until independently tested.

Full installation and troubleshooting guide. Check your teacher's or exam's rules before using any program.

Version history

1.0.0 — current version, published 2026-11-05. Download this reviewed version

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