总述:本次共解1题。其中
☆算术计算1题
〖1/1算式〗
作业:求算式 cos(arcsin(10/(4.8882+5.1118)))*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882) 的值.
题型:数学计算
解:
cos(arcsin(10/(4.8882+5.1118)))*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)
=cos(arcsin(10/10))*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)
=cos(arcsin1)*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)
=cos1.570796*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)
=cos1.570796*10-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)
=cos1.570796*10-cos(arcsin(10/0.2236))*(5.1118-4.8882)
=cos1.570796*10-cos(arcsin44.722719)*(5.1118-4.8882)
=cos1.570796*10-cos0*(5.1118-4.8882)
=cos1.570796*10-cos0*0.2236
=0*10-cos0.000000*0.2236
=0*10-1*0.2236
=0-1*0.2236
=0-0.2236
=-0.2236 答案:cos(arcsin(10/(4.8882+5.1118)))*(4.8882+5.1118)-cos(arcsin(10/(5.1118-4.8882)))*(5.1118-4.8882)=-0.2236 注:弧度制你的问题在这里没有得到解决?请到 热门难题 里面看看吧!