作业帮向量m=(2cosC/2,-sinC),向量n=(cosC/2,2sinC)且向量m⊥向量n.1求角C的大小.2若a^2
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向量m=(2cosC/2,-sinC),向量n=(cosC/2,2sinC)且向量m⊥向量n.1求角C的大小.2若a^2=2b^2+c^2,求tanA的值
m⊥n
=>m.n=0
(2cos(C/2),-sinC).(cos(C/2),2sinC)=0
2(cosC/2)^2-2(sinC)^2=0
(2(cosC/2)^2-1)-2(sinC)^2+1=0
cosC-2(sinC)^2+1=0
cosC - (2-2(cosC)^2)+1=0
2(cosC)^2+cosC-1=0
cosC = (-1+3)/4 = -1/2
C=2π/3
sinC = √3/2
a^2 =2b^2+c^2
c^2 = a^2+b^2 - 3b^2
by cosine rule
-3b^2 =-2abcosC
cosC = (3b/(2a)) = -1/2
b= -a/3
a^2 =2b^2+c^2
= b^2 + c^2 +b^2
by cosine -rule
-2bccosA = b^2
cosA = -b/(2c)
= -b/ (2asinC/sinA)
tanA = -2asinC/b
= -2a(√3/2))/( -a/3)
= 3√3
=>m.n=0
(2cos(C/2),-sinC).(cos(C/2),2sinC)=0
2(cosC/2)^2-2(sinC)^2=0
(2(cosC/2)^2-1)-2(sinC)^2+1=0
cosC-2(sinC)^2+1=0
cosC - (2-2(cosC)^2)+1=0
2(cosC)^2+cosC-1=0
cosC = (-1+3)/4 = -1/2
C=2π/3
sinC = √3/2
a^2 =2b^2+c^2
c^2 = a^2+b^2 - 3b^2
by cosine rule
-3b^2 =-2abcosC
cosC = (3b/(2a)) = -1/2
b= -a/3
a^2 =2b^2+c^2
= b^2 + c^2 +b^2
by cosine -rule
-2bccosA = b^2
cosA = -b/(2c)
= -b/ (2asinC/sinA)
tanA = -2asinC/b
= -2a(√3/2))/( -a/3)
= 3√3
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