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在三角形中,若8sin^2(B+C)/2-2cos2A=7,求角A.
人气:329 ℃ 时间:2020-05-05 15:39:55
解答
A+B+C=180 °
2A=360-(2B+2C)
cos2A=cos(2B+2C)=2[cos(B+C)]^2-1
∵cos2X=1-2(sinX)^2
∴(sinX)^2=(1-cos2X)/2
∴sin^2[(B+C)/2]=[1-cos(B+C)]/2
∴8sin^2[(B+C)/2]-2cos2A=7
4-4cos(B+C)-4[cos(B+C)]^2+2=7
4[cos(B+C)]^2+4cos(B+C)+1=0
[2cos(B+C)+1]^2=0
cos(B+C)=-1/2
∴B+C=120 °
∴A=60°
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