已知△ABC的三内角A、B、C满足A+C=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cos
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已知△ABC的三内角A、B、C满足A+C=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cosC)
(1)试求函数f(x)的解析式及定义域 (2)判断其单调性,并加以证明.
(1)试求函数f(x)的解析式及定义域 (2)判断其单调性,并加以证明.
1)
A+C=2B
B=π/3,cosB=1/2
f(x)=cosB(1/cosA+1/cosC)
=1/2*(cosA+cosC)/cosAcosC
=[cos(A+C)/2cos(A-C)/2] / [1/2*(cos(A+C)+cos(A-C))]
=[xcosB]/ [1/2*(cos2B+cos(A-C)]
=(x/2) / (1/2(-1/2+2cos^2(A-C)/2-1))
=x/(2x^2-3/2)
=2x/(4x^2-3)
定义域:x∈[0,1),x≠√3/2
2)
f(x)=2x/(4x^2-3)=2/(4x-3/x)
显然在定义域内是减函数
A+C=2B
B=π/3,cosB=1/2
f(x)=cosB(1/cosA+1/cosC)
=1/2*(cosA+cosC)/cosAcosC
=[cos(A+C)/2cos(A-C)/2] / [1/2*(cos(A+C)+cos(A-C))]
=[xcosB]/ [1/2*(cos2B+cos(A-C)]
=(x/2) / (1/2(-1/2+2cos^2(A-C)/2-1))
=x/(2x^2-3/2)
=2x/(4x^2-3)
定义域:x∈[0,1),x≠√3/2
2)
f(x)=2x/(4x^2-3)=2/(4x-3/x)
显然在定义域内是减函数
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