问题: 三角函数求值
x,y,z是锐角,
cosx+cosy+cosz -sin^x
=cos^x+4sin(x/2)sin(y/2)sin(z/2)
求x+y+z=?
解答:
cosx+cosy+cosz=1+4sin(x/2)sin(y/2)sin(z/2)
cosx+cosy=1-cosz+4sin(x/2)sin(y/2)sin(z/2)
2cos(x+y/2)cos(x-y)/2
=2sin^(z/2)-2sin(z/2)[cos(x+y)/2 -cos(x-y)/2]
[cos(x-y)/2 +sin(z/2)][cos(x+y)/2 -sin(z/2)]=0
x,y,z是锐角
显然[cos(x-y)/2 +sin(z/2)]>0
所以[cos(x+y)/2 -sin(z/2)]=0
<==>(x+y)/2 + z/2
==>x+y+z =18度0
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