数学题已知向量a=(√3,1),
是f(x)=2a*bsinx吧?如是,则:
(1):
f(x)=2(√3cosx-sinx)sinx
=2√3cosxsinx-2(sinx)^2
=√3sin(2x)+cos(2x)-1
=2sin(2x+π/6)-1
因x∈[0,π],则:π/6≤2x+π/6≤13π/6
故-1≤sin(2x+π/6)≤1
-3≤f(x)≤1
(2):
|a|=√((√3)^2+1^2)=2
|b|=√((cosx)^2+(-sinx)^2)=1
f(x)=1,则2sin(2x+π/6)-1=1,2x+π/6=π/2,x=π/6
b=(√3/2,-1/2)
则cosC=a*b/(|a||b|)=(3/...全部
是f(x)=2a*bsinx吧?如是,则:
(1):
f(x)=2(√3cosx-sinx)sinx
=2√3cosxsinx-2(sinx)^2
=√3sin(2x)+cos(2x)-1
=2sin(2x+π/6)-1
因x∈[0,π],则:π/6≤2x+π/6≤13π/6
故-1≤sin(2x+π/6)≤1
-3≤f(x)≤1
(2):
|a|=√((√3)^2+1^2)=2
|b|=√((cosx)^2+(-sinx)^2)=1
f(x)=1,则2sin(2x+π/6)-1=1,2x+π/6=π/2,x=π/6
b=(√3/2,-1/2)
则cosC=a*b/(|a||b|)=(3/2-1/2)/(2*1)=1/2
sinC=√(1-(cosC)^2)=√3/2
按余弦定理:
|c|^2=|a|^2+|b|^2-2|a||b|cosC
=4+1-2=3
|c|=√3
按正弦定理:
sinA=|a|*sinC/|c|=2*(√3/2)/√3=1
。
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