三角函数求值√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.我已经求得B=60°,A+C=120°,1/cosA-1/cos(60°+A)=-2√2.然后怎么办呢
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![三角函数求值√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.我已经求得B=60°,A+C=120°,1/cosA-1/cos(60°+A)=-2√2.然后怎么办呢](/uploads/image/z/8605660-4-0.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E6%B1%82%E5%80%BC%E2%88%9A%E8%A1%A8%E6%A0%B9%E5%8F%B7%2C%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E6%BB%A1%E8%B6%B3A%2BC%3D2B.%E4%B8%941%2FcosA%2B1%2FcosC%3D-%E2%88%9A2%2FcosB%2C%E6%B1%82cos%28A-C%29%2F2%E7%9A%84%E5%80%BC.%E6%88%91%E5%B7%B2%E7%BB%8F%E6%B1%82%E5%BE%97B%3D60%C2%B0%EF%BC%8CA%2BC%3D120%C2%B0%EF%BC%8C1%2FcosA-1%2Fcos%2860%C2%B0%2BA%29%3D-2%E2%88%9A2.%E7%84%B6%E5%90%8E%E6%80%8E%E4%B9%88%E5%8A%9E%E5%91%A2)
三角函数求值√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.我已经求得B=60°,A+C=120°,1/cosA-1/cos(60°+A)=-2√2.然后怎么办呢
三角函数求值
√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.
我已经求得B=60°,A+C=120°,1/cosA-1/cos(60°+A)=-2√2.然后怎么办呢
三角函数求值√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C)/2的值.我已经求得B=60°,A+C=120°,1/cosA-1/cos(60°+A)=-2√2.然后怎么办呢
因为:A+B+C=180,A+C=2B,
3B=180,B=60度
1/cosA+1/cosC=-√2/cosB,
(cosC+cosA)/cosA*cosC=-√2/cosB,
利用公式:
cosA+cosC=2*cos(A+C)/2*cos(A-C)/2,(和差化积)
cosA*cosC=1/2*[cos(A+C)+cos(A-C)],(积化和差公式)
就可得到:
[2*cos(A+C)/2*cos(A-C)/2]/{1/2*[cos(A+C)+cos(A-C)]=-√2/cosB,
[2cos60*cos(A-C)/2]*cosB=-√2*1/2*[cos120+cos(A-C)],
化简后,得
cos(A-C)/2=-√2*[cos(A-C)-1/2],
而,cos(A-C)=2cos^2(A-C)/2-1,(倍角公式),则有
2√2cos^2(A-C)/2+cos(A-C)/2-3√2/2=0,
令,cos(A-C)/2=t,|t|≤1,则有
2√2t^2+t-3√2/2=0,
4√2t^2+2t-3√2=0,
(2√2t+3)(2t-√2)=0,
t1=-3/(2√2),(不合,舍去,|t|≤1,)
t2=√2/2.
即,cos(A-C)/2=√2/2.
(注:三角函数主要是公式间的转化)