Question

一道关于高中数学三角函数题

已知在△ABC中，sinA(sinB+cosB)-sinC=0,sinB+cos2C=0,
求角A,B,C的大小。

Answer 1

sinC=sin(π-(A+B))=sin(A+B)
cos2C=cos2(π-(A+B))=cos2(A+B)
∴sinA(sinB+cosB)- sin(A+B)=0
sinAsinB+sinAcosB)- sinAcosB-cosAsinB=0
sinB(sinA-cosA)=0,又sinB≠0
∴sinA=cosA
∵A∈(0,π)
∴A= π/4
再由sinB+ cos2(A+B)=0
sinB+cos( π/2 +2B)=0
sinB-sin2B=0
cosB=- 1/2
∴B= π/3
又C=π-(A+B)=π-( π/3 + π/4 )= 5π/12
∴A= π/4 ,B= π/3 ,C= 5π/12 .

Answer 2

sinA(sinB+cosB)=sinC=sin(A+B)=sinAcosB+sinBcosA
得sinAsinB=sinBcosA
sinB不会等于0
所以sinA=cosA
所以A=45°
C=135°-B
sinB+cos2C=0
则sinB+cos(270°-2B)=0
即sinB+sin2B=0
sinB+2sinBcosB=0
cosB=-1/2
B=120°
则C=15°

Answer 3

sinA(sinB+cosB)-sinC=0 ----- sinB(sinA-cosA) = 0

sinA = cosA ,A = 45

sinB-cos2C= sinB - cos(2B + 90) = sinB - sin2B = sinB(1-2cosB) = 0

cosB = 1/2 , B = 60

C = ......

Answer 4

第一步：sinA(sinB+cosB)-sinC=0
展开 sinAsinB+sinAcosB=sincC
A+B+C=π