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(3/(2+10)))*(2+10)-cos(arcsin(3/(10-2)))*(10-2) 的值.
题型:数学计算
解:
cos(arcsin(3/(2+10)))*(2+10)-cos(arcsin(3/(10-2)))*(10-2)
=cos(arcsin(3/12))*(2+10)-cos(arcsin(3/(10-2)))*(10-2)
=cos(arcsin0.25)*(2+10)-cos(arcsin(3/(10-2)))*(10-2)
=cos0.25268*(2+10)-cos(arcsin(3/(10-2)))*(10-2)
=cos0.25268*12-cos(arcsin(3/(10-2)))*(10-2)
=cos0.25268*12-cos(arcsin(3/8))*(10-2)
=cos0.25268*12-cos(arcsin0.375)*(10-2)
=cos0.25268*12-cos0.384397*(10-2)
=cos0.25268*12-cos0.384397*8
=0.968246*12-cos0.384397*8
=0.968246*12-0.927025*8
=11.618952-0.927025*8
=11.618952-7.4162
=4.202752 答案:cos(arcsin(3/(2+10)))*(2+10)-cos(arcsin(3/(10-2)))*(10-2)=4.202752 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!