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The following program is suppose to use the sin() function in the math library and write out an input's absolute value over an interval. So for example if sine(0.6) is 0.564 then its absolute value is the same (0.564). But if sine(2.4) is -0.675 then its absolute value is 0.675. #include #include /* has sin(), abs(), and fabs() */ int main(void) { double interval; int i; for(i = 0; i <30; i++) { interval = i/10.0; printf("sin( %lf ) = %lf \t", interval, abs(sin(interval))); } printf("\n+++++++\n"); return 0; }

Sagot :

Answer:

#include <stdio.h>

#include <math.h> /* has sin(), abs(), and fabs() */

int main(void) {

   double interval;

   int i;

   for(i = 0; i <30; i++) {

       interval = i/10.0;

       printf("sin( %.1lf ) = %.3lf \t", interval, abs(sin(interval)));

   }

   printf("\n+++++++\n");

   return 0;

}

Explanation:

The C source code defines a void main program that outputs the absolute value of the sine() function of numbers 0.1 to 3.0.

MrRoyal

The program is an illustration of loops.

Loops are used to perform repetitive and iterative operations.

The program statements in C language that completes the program are as follows:

interval = i/10.0;

printf("sin( %.1lf ) = %.3lf \t", interval, abs(sin(interval)));

The flow of the above statements is as follows

  • The first line determines the interval
  • The second line prints the required output

Read more about similar programs at:

https://brainly.com/question/14575844

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