高中數學!急!三角函數!
題目:
高中數學!急!三角函數!
解答:
f(x)=1+sin2x+√3(1-cos2x)
=sin2x-√3cos2x+√3+1
=2sin(2x-π/3)+√3+1
1)T=2π/2=π
2)2acosC+c=2b
2sinAcosC+sinC=2sinB=2sin(A+C)=2sinAcosC+2cosAsinC
cosA=1/2
A=π/3
0
題目:
高中數學!急!三角函數!
解答:
f(x)=1+sin2x+√3(1-cos2x)
=sin2x-√3cos2x+√3+1
=2sin(2x-π/3)+√3+1
1)T=2π/2=π
2)2acosC+c=2b
2sinAcosC+sinC=2sinB=2sin(A+C)=2sinAcosC+2cosAsinC
cosA=1/2
A=π/3
0
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