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:

题目

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

答案

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