已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
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已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
A+C=2B,
B=60,A+C=120
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+cos(A-C)]
=-√2{2[cos(A-C)/2]^2-3/2}
cos[(A-C)/2]=t
t=-√2{2t^2-3/2}
t=√2/2
三角形ABC中,A+C=2B,
得:B=60, A+C=120
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC
(左边用和差化积,右边用积化和差)
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1...
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三角形ABC中,A+C=2B,
得:B=60, A+C=120
1/cosA+1/cosC=-√2/cosB =-2√2
cosA+cosC=-2√2cosAcosC
(左边用和差化积,右边用积化和差)
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+cos(A-C)]
=-√2{2[cos(A-C)/2]^2-3/2}
令:cos[(A-C)/2]=t
则:t=-√2{2t^2-3/2}
解得:t=√2/2
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已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
解 因为在△ABC中 所以 ∠A+∠B+∠C=180
3∠B=180 ∠B=60
1/cosA+1/cosC=-√2/cosB=-2√ 2=(cosA+c...
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已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
解 因为在△ABC中 所以 ∠A+∠B+∠C=180
3∠B=180 ∠B=60
1/cosA+1/cosC=-√2/cosB=-2√ 2=(cosA+cosC)/cosAcosC
根据公式 cos^2(α/2)=(1+cosα)/2
可得 cos(A-C)/2=√(1+cos(A-C))
cos(A-C)=cosAcosC+sinAsinC=a
cos(A+C)=cosAcosC-sinAsinC=-cosB=-1/2
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三角形ABC的三个内角A,B,C满足:A+C=2B,得B=60°,A+C=120°。
1/cosA+1/cosC=-√2/cosB,得cosA+cosC=-2√2cosA*cosC,
即2cos[(A-C)/2]cos[(A+C)/2]=(-√2)[cos(A+C)+cos(A-C)],
进而有cos[(A-C)/2]=(√2/2)-(√2)cos(A-C)
=(√...
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三角形ABC的三个内角A,B,C满足:A+C=2B,得B=60°,A+C=120°。
1/cosA+1/cosC=-√2/cosB,得cosA+cosC=-2√2cosA*cosC,
即2cos[(A-C)/2]cos[(A+C)/2]=(-√2)[cos(A+C)+cos(A-C)],
进而有cos[(A-C)/2]=(√2/2)-(√2)cos(A-C)
=(√2/2)-2√2cos^2[(A-C)/2]+(√2),
即2√2cos^2[(A-C)/2]+cos[(A-C)/2]-(3√2/2)=0,
cos(A-C)/2=√2/2。
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