Question

高三数学问题 谢谢大神

高三数学问题 谢谢大神图中第二问！！！！！方便的话用纸写！！谢谢

Answer 1

(1)
m.n= 2sinA.sinB

(sinA+sinB, sinC).(sinA+sinB,-sinC) =2sinA.sinB
(sinA+sinB)^2 - (sinC)^2 =2sinA.sinB
(a+b)^2 - c^2 = 2ab
a^2+b^2 -c^2 =0
c^2 = a^2+b^2
By cosine-rule
c^2 = a^2+b^2-2abcosC
=>
2abcosC = 0
C=π/2
(2)
S= sinA+sinB
= sinA + sin(A+C)
=sinA + cosA
S'= cosA -sinA
S'=0
cosA -sinA = 0
tanA =1
A =π/4
S'|A =π/4+ <0
S'|A =π/4- >0
A =π/4 (max)
max S = √2
S > sin0 + cos0 =1
1<sinA+sinB ≤√2

追问

那个C是60度的哦

追答

(1)
这样
m.n= 3sinA.sinB
(sinA+sinB, sinC).(sinA+sinB,-sinC) =3sinA.sinB
(sinA+sinB)^2 - (sinC)^2 =3sinA.sinB
(a+b)^2 - c^2 = 3ab
a^2+b^2 -c^2 =ab
c^2 = a^2+b^2 -ab
By cosine-rule
c^2 = a^2+b^2-2abcosC
=>
2abcosC = ab
cosC =1/2
C=π/3
(2)
S= sinA+sinB
= sinA + sin(A+C)
=sinA + sin(A+π/3)
=sinA + (1/2)sinA + (√3/2)cosA
=(3/2)sinA +(√3/2)cosA
S'= (3/2)cosA - (√3/2)sinA
S'=0
(3/2)cosA - (√3/2)sinA = 0
tanA = √3
A =π/3

S'|A =π/3+ 0
A =π/3 (max)
max S = sin(π/3) + sin(π/3+π/3) = √3
S > sin0 + sin(π/3)=√3/2
√3/2<sinA+sinB ≤√3

Answer 2

A+B=120度，，，，全部用A表示，A有个范围