c++:用下列泰勒级数求sinx的近似值,x的值从键盘输入,精度要求为10-6.sinx=x-(x3)/3!+(x5)/5!+.._其他解答

首先,我们知道sin(x) = sin(x+k*2pi),那么,我们首先去x1 = x%2pi.
这样做的目的,是为了让表达式在精度要求范围内,尽可能的有比较少的项.
然后循环累加,跳出循环的条件是最后一项的值小于10的-6次方.能给出具体的程式吗？？。。。大哥。。。你要不要这么懒啊。。。我刚学是个菜鸟#include using namespace std;const double min= (1.0e-6);int main(){ double x = 0; cout<<"input x = "; cin>>x; int index = 1; double p = x; double result = x; int flag = -1; while(p>min) {p = (p*x*x)/( (index+1)*(index+2) );result = result + flag*p;index = index + 2;flag = -flag; } cout<<"result is "<