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(2/(2+10)))*(2+10)-cos(arcsin(2/(10-2)))*(10-2) 的值.
题型:数学计算
解:
cos(arcsin(2/(2+10)))*(2+10)-cos(arcsin(2/(10-2)))*(10-2)
=cos(arcsin(2/12))*(2+10)-cos(arcsin(2/(10-2)))*(10-2)
=cos(arcsin0.166667)*(2+10)-cos(arcsin(2/(10-2)))*(10-2)
=cos0.167448*(2+10)-cos(arcsin(2/(10-2)))*(10-2)
=cos0.167448*12-cos(arcsin(2/(10-2)))*(10-2)
=cos0.167448*12-cos(arcsin(2/8))*(10-2)
=cos0.167448*12-cos(arcsin0.25)*(10-2)
=cos0.167448*12-cos0.25268*(10-2)
=cos0.167448*12-cos0.25268*8
=0.986013*12-cos0.252680*8
=0.986013*12-0.968246*8
=11.832156-0.968246*8
=11.832156-7.745968
=4.086188 答案:cos(arcsin(2/(2+10)))*(2+10)-cos(arcsin(2/(10-2)))*(10-2)=4.086188 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!