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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/20 03:00:28
![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](/uploads/image/z/8857002-66-2.jpg?t=f%28x%29%3D%281%2B1%2Ftanx%29sin%28x%29%5E2%2Bmsin%28x%2B45%29sin%28x-45%29%E5%BD%93tan%28a%29%3D2%E6%97%B6%2Cf%28a%29%3D3%2F5%2C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC)
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
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
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
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 = -3/5,从而 sin(a+45)sin(a-45) = -1/2cos2a = 3/10.
所以
3/5
=f(a)
=(1+1/tana)(sina)^2+msin(a+45)sin(a-45)
=(1+1/2)(4/5)+m(3/10)
由此可以解出 m = -2.
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/...
全部展开
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
收起
这么简单的题目 要是不会 就去问 坤哥啊