Question

一道高一数学题！！！~

Answer 1

答：
（3）设arcsin(4√3/7)=a，arcsin(1/7)=b
则：sina=4√3/7，sinb=1/7，sin²a+sin²b=1
所以：sina=cosb
所以：a+b=π/2
所以：arcsin(4√3/7)+arcsin(1/7)=π/2

（4）0<x<π/2
arcsin[cos(π/2+x)]+arccos[sin(π+x)]
=arcsin[cos(π-π/2+x)]+arccos[sin(π/2+π/2+x)]
=arcsin[-cos(π/2-x)]+arccos[cos(-π/2-x)]
=arcsin[sin(-x)]+arccos[cos(π/2+x)]
=-x+π/2+x
=π/2

（5）tan[2arctan(1/2)+arctan(1/3)]
={tan[2arctan(1/2)]+tan[arctan(1/3)]}/{1-tan[2arctan(1/2)]*tan[arctan(1/3)]}
={2*(1/2)/[1-(1/2)²]+1/3}/[1-(1/3)/(1/4)]
=(4+1/3)/(-1/3)
=-13