CODEWITHSUNDEEP
ccalculator
affanmansoor
affanmansoor

Reputation: 25

Sin function without using math.h library

I've got an assignment for FOP to make a scientific calculator, we haven't been taught about the math.h library! my basic approach for one of the function SIN was this but i'm failing to make this work

#include <stdio.h>
int main()
{
int input;
float pi;
double degree;
double sinx;
long int powerseven;
long int powerfive;
long int powerthree;
input = 5;
degree= (input*pi)/180;
pi=3.142;
powerseven=(degree*degree*degree*degree*degree*degree*degree);
powerfive=(degree*degree*degree*degree*degree);
powerthree=(degree*degree*degree);
sinx = (degree - (powerthree/6) + (powerfive/120) - (powerseven/5040));
printf("%ld", sinx);
getchar();
}

Upvotes: 0

Views: 6863

Answers (2)

R.. GitHub STOP HELPING ICE
R.. GitHub STOP HELPING ICE

Reputation: 215211

vacawama covered most of the technical C-language reasons your program isn't working. I'll attempt to cover some algorithmic ones. Using a fixed finite number of taylor series terms to compute sine is going to lose precision quickly as the argument gets farther away from the point at which you did the series expansion, i.e. zero.

To avoid this problem, you want to use the periodicity of the sine function to reduce your argument to a bounded interval. If your input is in radians, this is actually a difficult problem in itself, since pi is not representable in floating point. But as long as you're working in degrees, you can perform argument reduction by repeatedly subtracting the greatest power-of-two multiple of 360 that's less than the argument, until your result is in the interval [0,360). (If you could use the standard library, you could just use fmod for this.)

Once your argument is in a bounded interval, you can just choose an approximation that's sufficiently precise on that interval. A taylor series approximation is certainly one approach you can use at this point, but not the only one.

Upvotes: 3

vacawama
vacawama

Reputation: 154583

Your code almost works. You have a few problems:

  1. You are using pi before initializing it. I suggest using a more accurate value of pi such as 3.14159265359.

  2. powerseven, powerfive and powerthree should be defined as double instead of as long int. You are losing precision by storing these values in an integer type. Also, when you divide an integer value by an integer value (such as powerthree/6) the remainder is lost. For instance, 9/6 is 1.

  3. Since sinx is a double you should be using printf("%f", sinx);

Upvotes: 4

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