已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
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![已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长](/uploads/image/z/5257571-59-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%283pai%2F2-x%29cosx-sinxcos%28pai%2Bx%29%281%29%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4.%282%29%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92A.B.C%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E8%8B%A5A%E4%B8%BA%E9%94%90%E8%A7%92%2Cf%28A%29%3D0%2CBC%3D2%2C%E6%B1%82AC%E7%9A%84%E9%95%BF)
已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
已知函数f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)(1)求函数的单调递增区间.(2)三角形ABC的三个内角A.B.C成等差数列,若A为锐角,f(A)=0,BC=2,求AC的长
f(x)=sin(3pai/2-x)cosx-sinxcos(pai+x)
=-cosx*cosx-sinx*(-cosx)
=-(1+cos2x)/2+(1/2)sin2x
=(1/2)sin2x-(1/2)cos2x-1/2
=(√2/2)*[sin2x*cos(π/4)-cos2x*sin(π/4)]-1/2
=(√2/2)sin(2x-π/4)-1/2
(1)增区间
2kπ-π/2≤2x-π/4≤2kπ+π/2
2kπ-π/4≤2x≤2kπ+3π/4
即 kπ-π/8≤x≤kπ+3π/8
∴ 增区间为[ kπ-π/8,kπ+3π/8],k∈Z
(2)三角形ABC的三个内角A.B.C成等差数列,
则A+C=2B
∴ 3B=A+B+C=π
∴ B=π/3
f(A)=0,∴ (√2/2)sin(2A-π/4)-1/2=0
∴ sin(2A-π/4)=√2/2
∴ A=π/4
利用正弦定理AC/sinB=BC/sinA
∴ AC=BCsinB/sinA=2*(√3/2)/(√2/2)=√6
先化为一体,再求