输入精度e 和实数x,用下列公式求cos x 的近似值,精确到最后一项的绝对值...答:{ int i;double m=1.0;for(i=1;i<=n;i++)m*=i;return m;} double funcos(double e,double x){ int i=0,flag=1;double cos=0,item=1.0;while(fabs(item)>=e){ item=flag*pow(x,i)/f(i);cos+=item;flag=-flag;//printf("%lf %lf %lf\n",pow(x,i),f(i),item...