C (gcc), 178 172 bytes
double d;_;f(double(*x)(double)){d=x(0.9247);_=*(int*)&d%12;puts((char*[]){"acosh","sinh","asinh","atanh","tan","cosh","asin","sin","cos","atan","tanh","acos"}[_<0?-_:_]);}
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Old but cool: C (gcc), 194 bytes
double d;_;f(double(*x)(double)){char n[]="asinhacoshatanh";d=x(0.9247);_=*(int*)&d%12;_=(_<0?-_:_);n[(int[]){10,5,5,0,14,10,4,4,9,14,0,9}[_]]=0;puts(n+(int[]){5,1,0,10,11,6,0,1,6,10,11,5}[_]);}
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The -lm switch in TIO is merely to test. If you could write a perfect
implementation of the standard trig functions you would get the right answer.
Explanation
The idea was to find some input value such that when I interpret the outputs of
each of the trig functions as integers they have different remainders modulo 12.
This will allow them to be used as array indices.
In order to find such an input value I wrote the following snippet:
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
// Names of trig functions
char *names[12] = {"sin","cos","tan","asin","acos","atan","sinh","cosh","tanh","asinh","acosh","atanh"};
// Pre-computed values of trig functions
double data[12] = {0};
#define ABS(X) ((X) > 0 ? (X) : -(X))
// Performs the "interpret as abs int and modulo by" operation on x and i
int tmod(double x, int i) {
return ABS((*(int*)&x)%i);
}
// Tests whether m produces unique divisors of each trig function
// If it does, it returns m, otherwise it returns -1
int test(int m) {
int i,j;
int h[12] = {0}; // stores the modulos
// Load the values
for (i = 0; i < 12; ++i)
h[i] = tmod(data[i],m);
// Check for duplicates
for (i = 0; i < 12; ++i)
for (j = 0; j < i; ++j)
if (h[i] == h[j])
return -1;
return m;
}
// Prints a nicely formatted table of results
#define TEST(val,i) printf("Value: %9f\n\tsin \tcos \ttan \n \t%9f\t%9f\t%9f\na \t%9f\t%9f\t%9f\n h\t%9f\t%9f\t%9f\nah\t%9f\t%9f\t%9f\n\n\tsin \tcos \ttan \n \t%9d\t%9d\t%9d\na \t%9d\t%9d\t%9d\n h\t%9d\t%9d\t%9d\nah\t%9d\t%9d\t%9d\n\n",\
val,\
sin(val), cos(val), tan(val), \
asin(val), acos(val), atan(val),\
sinh(val), cosh(val), tanh(val),\
asinh(val), acosh(val), atanh(val),\
tmod(sin(val),i), tmod(cos(val),i), tmod(tan(val),i), \
tmod(asin(val),i), tmod(acos(val),i), tmod(atan(val),i),\
tmod(sinh(val),i), tmod(cosh(val),i), tmod(tanh(val),i),\
tmod(asinh(val),i), tmod(acosh(val),i), tmod(atanh(val),i))
// Initializes the data array to the trig functions evaluated at val
void initdata(double val) {
data[0] = sin(val);
data[1] = cos(val);
data[2] = tan(val);
data[3] = asin(val);
data[4] = acos(val);
data[5] = atan(val);
data[6] = sinh(val);
data[7] = cosh(val);
data[8] = tanh(val);
data[9] = asinh(val);
data[10] = acosh(val);
data[11] = atanh(val);
}
int main(int argc, char *argv[]) {
srand(time(0));
// Loop until we only get 0->11
for (;;) {
// Generate a random double near 1.0 but less than it
// (experimentally this produced good results)
double val = 1.0 - ((double)(((rand()%1000)+1)))/10000.0;
initdata(val);
int i = 0;
int m;
// Find the smallest m that works
do {
m = test(++i);
} while (m < 0 && i < 15);
// We got there!
if (m == 12) {
TEST(val,m);
break;
}
}
return 0;
}
If you run that (which needs to be compiled with -lm) it will spit out that with
a value of 0.9247 you get unique values.
Next I reinterpeted as integers, applied modulo by 12, and took the absolute
value. This gave each function an index. They were (from 0 -> 11): acosh, sinh,
asinh, atanh, tan, cosh, asin, sin, cos, atan, tanh, acos.
Now I could just index into an array of strings, but the names are very long and
very similar, so instead I take them out of slices of a string.
To do this I construct the string "asinhacoshatanh" and two arrays. The first
array indicates which character in the string to set to the null terminator,
while the second indicates which character in the string should be the first
one. These arrays contain: 10,5,5,0,14,10,4,4,9,14,0,9 and
5,1,0,10,11,6,0,1,6,10,11,5 respectively.
Finally it was just a matter of implementing the reinterpretation algorithm
efficiently in C. Sadly I had to use the double type, and with exactly 3 uses,
it was quicker to just use double three times then to use #define D double\nDDD
by just 2 characters. The result is above, a description is below:
double d;_; // declare d as a double and _ as an int
f(double(*x)(double)){ // f takes a function from double to double
char n[]="asinhacoshatanh"; // n is the string we will manipulate
int a[]={10,5,5,0,14,10,4,4,9,14,0,9}; // a is the truncation index
int b[]={5,1,0,10,11,6,0,1,6,10,11,5}; // b is the start index
d=x(0.9247); // d is the value of x at 0.9247
_=*(int*)&d%12; // _ is the remainder of reinterpreting d as an int and dividing by 12
_=(_<0?-_:_); // make _ non-negative
n[a[_]]=0; // truncate the string
puts(n+b[_]);} // print the string starting from the correct location
Edit: Unfortunately just using a raw array is actually shorter, so the code becomes much simpler. Nonetheless the string slicing was fun. In theory an appropriate argument might actually come up with the right slices on its own with some math.
1Mathematica can handle symbolic inputs so that the function output is only partially evaluated, if at all. The difference it makes is that I could use some pattern-matching instead of computations. – JungHwan Min – 2018-07-10T18:26:59.863
1@JungHwanMin If that means you can access the function names from the symbolic output then I'm afraid it is not allowed. – Laikoni – 2018-07-10T18:39:47.673