f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C
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![f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C](/uploads/image/z/1856251-19-1.jpg?t=f%28x%29%3D2sin%282x%2B2A%2B%CF%80%2F6%29%2B1%2CA%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%EF%BC%881%EF%BC%89%E8%AE%BE%E2%88%A0A%2C%E2%88%A0B%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E8%A7%92%2C%E8%8B%A5f%28B%29%3D0%2C%E6%B1%82%E2%88%A0C)
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C
f(B)=2sin(2B+2A+π/6)+1=0
sin(2B+2A+π/6)=-1/2=sin7π/6或sin11π/6
2A+2B+π/6=7π/6或11π/6
A+B=π/2或5π/6
C=π-(A+B)=π/2或π/6
∠A,∠B是三角形ABC的内角,所以sin(A+B)=sinC,cos(A+B)=-cosC
f(B)=2sin(2B+2A+π/6)+1
=√3sin2(A+B)+cos2(A+B)+1
=2√3sin(A+B)cos(A+B)+2[cos(A+B)]^2
=-2√3sinC*cosC+2[cosC]^2
=2cosC*(cosC-√3sinC)
=0
所以cosC=0或tanC=√3/3
C=π/2或C=π/6
f(x)=2sin(2x+2A+π/6)+1=2[sin2(x+A)cos(π/6)+cos2(x+A)sin(π/6)]+1=√3sin2(x+A)+cos2(x+A)+1
设∠A,∠B是三角形ABC的内角,若f(B)==√3sin2(B+A)+cos2(B+A)+1=√3sin2(π-C)+cos2(π-C)+1=0
即f(B)=2sin[(π/6-2C)+1=2sin[(π/...
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f(x)=2sin(2x+2A+π/6)+1=2[sin2(x+A)cos(π/6)+cos2(x+A)sin(π/6)]+1=√3sin2(x+A)+cos2(x+A)+1
设∠A,∠B是三角形ABC的内角,若f(B)==√3sin2(B+A)+cos2(B+A)+1=√3sin2(π-C)+cos2(π-C)+1=0
即f(B)=2sin[(π/6-2C)+1=2sin[(π/6-2C)+2π]+1=2sin(π/6-2C)+1=0
2sin(π/6-2C)=-1=sin(-π/2)
π/6-2C=-π/2
所以∠C=π/3
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