用C语言计算余弦函数的近似值

用c语言余弦函数的近似值

不知道题目的目的是考读写格式还是计算。假定是计算，程序如下：

#include

#include

#include

double funcos(double e,double x){

double sum=1.0,term=1;

int i2=0,x2;

while(1){

term= (-1) * term*x/(i2+1) *x /(i2+2);

sum=sum+term;

i2=i2+2;

if (fabs(term) < e) break;

}

return sum;

}

main(){

double e,x,v;

printf(＂input e: ＂);

scanf(＂%lf＂,&e);

printf(＂ninput x: ＂);

scanf(＂%lf＂,&x);

v = funcos(e,x);

printf(＂ncos(x)=%.3lfn＂,v);

return 0;

}

========

如果是考输入

e: 0.001x: 1

的读语句，那么上面程序修改一下：

printf(＂inputne: 0.001x: 1n＂);

scanf(＂e: %lf x:%lf＂,&e,&x);

要考输出花样，更简单，你自己能写，就不多说了。

C语言：输入精度e和实数x用下列公式cos x的近似值精确到最

#include

#include

double funcos(double x,double e);

int main(void)

{

double n,x,e;

scanf(＂e: %lf%*c＂,&e);

scanf(＂x: %lf%*c＂,&x);

n=funcos(x,e);

printf(＂cos(x)=%.3lf＂,n);

return 0;

}

double funcos(double x,double e){

int i,k,flag=1;

double sum,result=0,fact=1;

for(i=0;;i+=2){

sum=x ;

fact=1;

for(k=1;k<=i;k++)

{

sum *=x ;

fact=fact*k;

}

sum/=fact;

result=result+flag*sum;

if(sum

break;

flag=-flag;

}

return result;

}

C语言编程逻辑问题

#include

using namespace std;

double funCos(double e, float x)

{

int i, m, n, p;

i = 2; m = x; n = 1; p = -1;

double s = 1.0, sum;

sum = 1;

while (s >e)

{

m = 1;

for (int j = 1; j {

m = m*x;

}

n = 1;

for (int k = 1; k {

n = n*k;

}

sum = sum + p*(m / n);

s = (m / n);

p = p*(-1);

i = i + 2;

}

return sum;

}

void main()

{

double e, t;

int x;

cout cin >>x >>e;

t = funCos(e, x);

cout }

c编写程序用公式计算e的近似值。直到最后一项小于输入的精度

//利用公式e=1+1/1!+1/2!+...+1/n!

#include ＂iostream＂

using namespace std;

double fact(int n)

{

double sum=1.0;

int i;

for(i=1;i<=n;i++)

sum*=i;

return sum;

}

int main()

{

double e=0,f,item=1;

int i;

cout<<；＂请输入精度：＂；

cin>>f;

for(i=0;;i++)

{

item=1/fact(i);

e+=item;

if(item

break;

}

cout<<＂e=＂<

}