△ABC中,∠A,B,C所对的边为a,b,c,且满足cos2A-cos2B=2cos(π/6-A)cos(π/6+A) 1.求∠B
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![△ABC中,∠A,B,C所对的边为a,b,c,且满足cos2A-cos2B=2cos(π/6-A)cos(π/6+A) 1.求∠B](/uploads/image/z/1038767-23-7.jpg?t=%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0A%2CB%2CC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E4%B8%BAa%2Cb%2Cc%2C%E4%B8%94%E6%BB%A1%E8%B6%B3cos2A-cos2B%3D2cos%28%CF%80%2F6-A%EF%BC%89cos%28%CF%80%2F6%2BA%EF%BC%89+1.%E6%B1%82%E2%88%A0B)
△ABC中,∠A,B,C所对的边为a,b,c,且满足cos2A-cos2B=2cos(π/6-A)cos(π/6+A) 1.求∠B
△ABC中,∠A,B,C所对的边为a,b,c,且满足cos2A-cos2B=2cos(π/6-A)cos(π/6+A) 1.求∠B
△ABC中,∠A,B,C所对的边为a,b,c,且满足cos2A-cos2B=2cos(π/6-A)cos(π/6+A) 1.求∠B
∵2cos(π/6-A)cos(π/6+A)=2(√3/2*cosA+1/2*sinA)(√3/2*cosA-1/2*sinA)
=2(3/4cos²A-1/4*sin²A)=3/2cos²A-1/2*sin²A
∵cos2A-cos2B=1-2sin²A-(2cos²B-1)=2-2sin²A-2cos²B
∴3/2cos²A-1/2*sin²A=2-2sin²A-2cos²B
3/2cos²A+3/2*sin²A=2-2cos²B
3/2=2-2cos²B
2cos²B=1/2,cos²B=1/4
cosB=±1/2
所以B=120或B=60
作参考吧
2cos(Pai/6-A)cos(Pai/6+A)=2[根号3/2cosA-1/2sinA][根号3/2cosA+1/2sinA]
=2[3/4cos^2A-1/4sin^2A]
cos2A-cos2B=2cos^2A-1-(2cos^2B-1)=2cos^2A-2cos^2 B=3/2cos^2A-1/2sin^2A
-2cos^2B=-1/2(cos^2A+sin^2A)
即有cos^2B=1/4
cosB=(+/-)1/2
故B=60或120度.
cos2A-cos2B=2cos(π/6-A)cos(π/6+A)
cos2A-cos2B=2(cos^2 A-cos^2 B)
角B=60°