在三角形ABC中,abc分别是ABC的对边,S为三角形的面积 若向量p=(4,a^2+b^2-c^2...

(1)
cosC=(a^2+b^2-c^2)/2ab
sinC = √[1- ((a^2+b^2-c^2)/2ab)^2]
p//q
=> 4/(a^2+b^2-c^2) =√3/S
S=√3(a^2+b^2-c^2)/4
(1/2)absinC =√3(a^2+b^2-c^2)/4
(1/2)ab√[1- ((a^2+b^2-c^2)/2ab)^2] = √3(a^2+b^2-c^2)/4
√[1- ((a^2+b^2-c^2)/2ab)^2]=√3(a^2+b^2-c^2)/ (2ab)
1- ((a^2+b^2-c^2)/2ab)^2 = 3[(a^2+b^2-c^2)/ (2ab)]^2
4[(a^2+b^2-c^2)/ (2ab)]^2=1
(a^2+b^2-c^2)/ (2ab)=1/2
(a^2+b^2-c^2)=ab
c^2=a^2+b^2-ab
by cosine rule
-ab =-2abcosC
cosC = 1/2
C =π/3
(2)
f(x)=4sinxcos(x+π/6)+1
f'(x) =4(-sinxsin(x+π/6) +cosxcos(x+π/6) ) =0
=>cos(2x+π/6) =0
2x+π/6 = π/2
x= π/6
ie A =π/6
f(π/6) = 4sin(π/6)cos(π/3)+1=2 =b
A=π/6 ,C=π/3 ,=> B= π/2
by sine rule
a/sinA =b/sinB
a = bsinA = 2sin(π/6)=1
c=bsinC =2(√3/2)=√3
S = (1/2)ac = √3/2

三角形

因为p平行于q，所以4(½absinC)=根号3(a²+b²+c²)解得tanC=根号3，所以，角C等于60度

(1)p∥q 4s=(a^2+b^2-c^2)*√3 (a^2+b^2-c^2)=(4s)/√3
cosC=[(4s)/√3]/2ab=[2absinC]/√3*2ab=sinC/√3
so tanC=√3 C=π/3
(2)f(A)=4sinAcos(A+π/6)+1=4sinAcosA*√3/2-4sin^2A*1/2+1=√3sin2A-(1-cos2A)+1
=2sin(2A+π/6)
b=2 A=π/6 C=π/3 B=π/2
so a=1 b=√3 S=1/2*1*2=1