Question

求化简这个式子

Answer 1

cos [ arcsin(3/5) + arccos (-12/13) ]
=cos[arcsin(3/5)].cos[arccos (-12/13)] -sin[arcsin(3/5)].sin[arccos (-12/13)]
=cos[arccos(4/5)].cos[arccos (-12/13)] -sin[arcsin(3/5)].sin[arcsin (-5/13)]
=(4/5)(-12/13) - (3/5)(-5/13)
=(-48+15)/65
=-33/65

追问

感谢解答！但是答案好像和楼下有区别

追答

错了一个正负号
cos [ arcsin(3/5) + arccos (-12/13) ]
=cos[arcsin(3/5)].cos[arccos (-12/13)] -sin[arcsin(3/5)].sin[arccos (-12/13)]
=cos[arccos(4/5)].cos[arccos (-12/13)] -sin[arcsin(3/5)].sin[arcsin (5/13)]
=(4/5)(-12/13) - (3/5)(5/13)
=(-48-15)/65
=-63/65

Answer 2

设a=arcsin(3/5)
则sina=3/5
且0<a<π/2
所以cosa>0
由sin²a+cos²a=1
所以cosa=4/5

令b=arccos(-12/13)
则cosb=-12/13
且π/2<b<π
所以sinb>0
由sin²b+cos²b=1
sinb=5/13

所以原式=cos(a+b)
=cosacosb-sinasinb
=-63/65

追问

你好，谢谢解答！不过你的答案好像和楼上不一样？

追答

连我的回答都不信？
采纳吧

Answer 3

八年级下册预备班