已知sin的平方2a+sin2acosa-cos2a=1,a属于(0,pai/2),(1)求tana的值;(2)求使函数
3个回答
(1) sin的平方2a+sin2acosa-cos2a=1
sin2acosa-cos2a=1-(sin2a)^2=(cos2a)^2
sin2acosa=cos2a(1+cos2a)
2sina(cosa)^2=cos2a*2(cosa)^2
sina=cos2a=1-2(sina)^2
2(sina)^2+sina-1
sina=-1,sina=1/2
∵a∈(0,π/2)
∴sina=1/2,a=π/6
tana=tan(π/6)=√3/3
(2)f(x)=sin的平方(x+a)+sin(a+x)cos(a+x)
=1/2[sin(2x+2a)-cos(2x+2a)+1]
=√2/2[sin(2x+2*π/6)cos(π/4)-cos(2x+2*π/6)sin(π/4)]+1/2
=√2/2sin(2x+2*π/6-π/4)+1/2
=√2/2sin(2x+π/12)+1/2
sin(2x+π/12)=1时,f(x)取得最大值
2x+π/12=2kπ+π/2
x=kπ+5π/24
因此,{x|x=kπ+5π/24}
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