Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?还是正向推理吧!
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 02:11:17
![Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?还是正向推理吧!](/uploads/image/z/5411056-40-6.jpg?t=Sin%281%2Bh%29-Sin1%3D2Sin%28h%2F2%29%2ACos%281%2Bh%2F2%29%E8%BF%99%E4%B8%AA%E6%88%90%E7%AB%8B%E5%90%97%3F%E8%BF%98%E6%98%AF%E6%AD%A3%E5%90%91%E6%8E%A8%E7%90%86%E5%90%A7%EF%BC%81)
Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?还是正向推理吧!
Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?
还是正向推理吧!
Sin(1+h)-Sin1=2Sin(h/2)*Cos(1+h/2)这个成立吗?还是正向推理吧!
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
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。