已知函数f (x )=2cosxsin(x-π/6)-1/2 求函数f(x)的最小值和最小正周期 是△ABC的内角A,B,C的对边分别为a,b,c且c=根号3,角C满足f(C)=0,若sinB=2sinA,求a,b的值
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![已知函数f (x )=2cosxsin(x-π/6)-1/2 求函数f(x)的最小值和最小正周期 是△ABC的内角A,B,C的对边分别为a,b,c且c=根号3,角C满足f(C)=0,若sinB=2sinA,求a,b的值](/uploads/image/z/3926077-61-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f+%28x+%29%3D2cosxsin%28x-%CF%80%2F6%29-1%2F2+%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F+%E6%98%AF%E2%96%B3ABC%E7%9A%84%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%E4%B8%94c%3D%E6%A0%B9%E5%8F%B73%2C%E8%A7%92C%E6%BB%A1%E8%B6%B3f%28C%29%3D0%2C%E8%8B%A5sinB%3D2sinA%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC)
已知函数f (x )=2cosxsin(x-π/6)-1/2 求函数f(x)的最小值和最小正周期 是△ABC的内角A,B,C的对边分别为a,b,c且c=根号3,角C满足f(C)=0,若sinB=2sinA,求a,b的值
已知函数f (x )=2cosxsin(x-π/6)-1/2 求函数f(x)的最小值和最小正周期 是△ABC的内角A,B,C的对边分别
为a,b,c且c=根号3,角C满足f(C)=0,若sinB=2sinA,求a,b的值
已知函数f (x )=2cosxsin(x-π/6)-1/2 求函数f(x)的最小值和最小正周期 是△ABC的内角A,B,C的对边分别为a,b,c且c=根号3,角C满足f(C)=0,若sinB=2sinA,求a,b的值
f (x )=2cosxsin(x-π/6)-1/2=sin(2x-π/6)-1, 最小值=-2,最小正周期=π. f(C)=sin(2C-π/6)-1=0,sin(2C-π/6)=1,-π/6<2C-π/6<11π/6, ∴2C-π/6=π/2,C=π/3. 由sinB=2sinA得b=2a, 由余弦定理,3=5a^-2a^,a^=1,a=1,b=2.
f(x)=2cosxsin(x-π/6)-1/2
=2(√3/2cosxsinx-1/2cos²x)-1/2
=√3/2sin2x-(1-cos2x)/2-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
所以f(x)的最小值为-2,最小正周期为π
2)c=√3 ...
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f(x)=2cosxsin(x-π/6)-1/2
=2(√3/2cosxsinx-1/2cos²x)-1/2
=√3/2sin2x-(1-cos2x)/2-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
所以f(x)的最小值为-2,最小正周期为π
2)c=√3 f(C)=sin(2C-π/6)-1=0
所以sin(2C-π/6)=1
即(2C-π/6)=π/2 (因为角C在三角形中,故不考虑其他周期的)
C=π/3
sinB=2sinA 即b=2a
cosC=(a²+b²-c²)/2ab=(5a²-3)/4a²=1/2
解得:a=1 b=2
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已知函数f (x )=2cosxsin(x-π/6)-1/2 ;(1)求函数f(x)的最小值和最小正周期;(2) △ABC的内角A,B,C的对边分别为a,b,c且c=√3,角C满足f(C)=0,若sinB=2sinA,求a,b的值;
(1)。f(x)=2cosx[sinxcos(π/6)-cosxsin(π/6)]-1/2
=2cosx[(√3/2)sinx-(1/2)cosx]-...
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已知函数f (x )=2cosxsin(x-π/6)-1/2 ;(1)求函数f(x)的最小值和最小正周期;(2) △ABC的内角A,B,C的对边分别为a,b,c且c=√3,角C满足f(C)=0,若sinB=2sinA,求a,b的值;
(1)。f(x)=2cosx[sinxcos(π/6)-cosxsin(π/6)]-1/2
=2cosx[(√3/2)sinx-(1/2)cosx]-1/2
=(√3)sinxcosx-cos²x-1/2=(√3/2)sin2x-(1+cos2x)/2-1/2
=(√3/2)sin2x-(1/2)cos2x-1=sin2xcos(π/6)-cos2xsin(π/6)-1=sin(2x-π/6)-1;故-2≦f(x)≦0;
即最小值为-2;最小正周期为π.
(2)。由sinB=2sinA,得sinA/sinB=a/b=2,故a=2b;c=√3;
f(C)=sin(2C-π/6)-1=0,故2C-π/6=π/2,故C=π/3;
由余弦定理。得3=a²+b²-2abcosC=4b²+b²-2b²=3b²,
故得b²=1,b=1,a=2.
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