f(x)=(1+1/tanx)sin(x)^2+msin(x+45)sin(x-45)当tan(a)=2时,f(a)=3/5,求m的取值
问题描述：
f(x)=(1+1/tanx)sin(x)^2+msin(x+45)sin(x-45)
当tan(a)=2时,f(a)=3/5,求m的取值
答
这么简单的题目 要是不会 就去问 坤哥啊
答
f(x)=(1+1/tanx)sin(x)^2+msin(x+45)sin(x-45)
=（1+1/tanx)sin(x)^2+msin(x+45)[-cos(x+45)]
=(1+1/tanx)sin(x)^2-msin(x+45)cos(x+45)
=(1+1/tanx)sin(x)^2-m/2*sin(2x+90)
=(1+1/tanx)sin(x)^2-m/2*cos2x
=3/2sin²x+msin²x-m/2
=（3/2+m）sin²x-m/2
tana=2 sin²a=4/5
f（a）=3/5
解得m=-2
答
当 tana=2 时,sina/cosa=2,(sina)^2/(cosa)^2=4,因此由 (sina)^2+(cosa)^2=1 即知 (sina)^2=4/5.由积化和差公式：sin(x+45)sin(x-45) = -1/2cos2x.由上述,tana=2,(sina)^2=4/5,所以由倍角公式可知 cos2a=1-2(sina)^2...