Question

在三角形ABC中，内角ABC的边长分别为abc，且满足A+C=3B，cos（B+C）=-3/5 1:

在三角形ABC中，内角ABC的边长分别为abc，且满足A+C=3B，cos（B+C）=-3/5
1:求sinC的值
2若a=5，求三角形面积
另外sin2A与sinA是什么关系，怎么转化

Answer 1

cosA = cos(π-(B+C)) = -cos(B+C) = 3/5
sinA = √(1-(3/5)² = 4/5

A+B+C= B+3B =π
B=π/4

1
sinC = sin(π-(A+B)) = sin(A+B)
= sinAcosB + cosAsinB
= 4/5 * √2/2 + 3/5 * √2/2
= √2/2 * (4/5 +3/5)
= 7√2/10

2
正弦定理
a/sinA = b/sinB =k
k = a/sinA =5/(4/5)= 25/4

b = k sinB = 25/4 * √2/2 = 25√2/8
三角形面积= 1/2 a*b *sinC = 1/2 * 5 * 25√2/8 * 7√2/10
= 25*7/16
= 175/16

3
sin2A = 2sinAcosA

Answer 2

（B+C）=180-A
cos（B+C）=-cosA=-3/5
cosA=3/5
sinA=4/5

满足A+C=3B
180-B=3B B=45
cos（45+C）=-3/5
sin（45+C）=4/5
cos45cosC-sin45sinC= -3/5
sin45cosC+cos45sinC= 4/5
sinC=7√2/10

a/sinA=b/sinB
bsinA=a*sinB b=25√2/8
面积S=1/2*absinC=1/2*5*25√2/8*7√2/10＝175/16
cosA=3/5
sinA=4/5

sin(2A)=2sinAcosA=24/25