Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?还是正向推理吧！

2Sin(h/2)*Cos(1+h/2)
=2Sin(h/2)*[cos1*cos(h/2)-sin1*sin(h/2)]
=sinh*cos1-(1-cosh)*sin1
=sinh*cos1+cosh*sin1-sin1
=sin(1+h)-sin1
故原式子成立。

1+h=1+h/2+h/2
1 =1+h/2-h/2
sin(1+h)=sin(1+h/2)*cos(h/2)+sin(h/2)*cos(1+h/2)
sin(1) =sin(1+h/2)*cos(h/2)-sin(h/2)*cos(1+h/2)
上面两式相减