在△ABC中,已知sin[B+(C/2)]=4/5,求cos(A-B)的值.【过程中有一步不懂,cos(A-B)=cos{[A+(C/2)]-[B+(C/2)]}=cos[A+(C/2)]cos[B+(C/2)]+sin[A+(C/2)]sin[B+(C/2)]∵sin[B+(C/2)]=4/5∵A+B+C=π∴A+(C/2)=π-B-(C/2)∴sin[A+(C/2)]=4/5∴A+(C/2)≠
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![在△ABC中,已知sin[B+(C/2)]=4/5,求cos(A-B)的值.【过程中有一步不懂,cos(A-B)=cos{[A+(C/2)]-[B+(C/2)]}=cos[A+(C/2)]cos[B+(C/2)]+sin[A+(C/2)]sin[B+(C/2)]∵sin[B+(C/2)]=4/5∵A+B+C=π∴A+(C/2)=π-B-(C/2)∴sin[A+(C/2)]=4/5∴A+(C/2)≠](/uploads/image/z/14319861-69-1.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5sin%5BB%2B%28C%2F2%29%5D%3D4%2F5%2C%E6%B1%82cos%28A-B%29%E7%9A%84%E5%80%BC.%E3%80%90%E8%BF%87%E7%A8%8B%E4%B8%AD%E6%9C%89%E4%B8%80%E6%AD%A5%E4%B8%8D%E6%87%82%2Ccos%28A-B%29%3Dcos%7B%5BA%2B%28C%2F2%29%5D-%5BB%2B%28C%2F2%29%5D%7D%3Dcos%5BA%2B%28C%2F2%29%5Dcos%5BB%2B%28C%2F2%29%5D%2Bsin%5BA%2B%28C%2F2%29%5Dsin%5BB%2B%28C%2F2%29%5D%E2%88%B5sin%5BB%2B%28C%2F2%29%5D%3D4%2F5%E2%88%B5A%2BB%2BC%3D%CF%80%E2%88%B4A%2B%28C%2F2%29%3D%CF%80-B-%28C%2F2%29%E2%88%B4sin%5BA%2B%28C%2F2%29%5D%3D4%2F5%E2%88%B4A%2B%28C%2F2%EF%BC%89%E2%89%A0)
在△ABC中,已知sin[B+(C/2)]=4/5,求cos(A-B)的值.【过程中有一步不懂,cos(A-B)=cos{[A+(C/2)]-[B+(C/2)]}=cos[A+(C/2)]cos[B+(C/2)]+sin[A+(C/2)]sin[B+(C/2)]∵sin[B+(C/2)]=4/5∵A+B+C=π∴A+(C/2)=π-B-(C/2)∴sin[A+(C/2)]=4/5∴A+(C/2)≠
在△ABC中,已知sin[B+(C/2)]=4/5,求cos(A-B)的值.【过程中有一步不懂,
cos(A-B)=cos{[A+(C/2)]-[B+(C/2)]}=cos[A+(C/2)]cos[B+(C/2)]+sin[A+(C/2)]sin[B+(C/2)]
∵sin[B+(C/2)]=4/5∵A+B+C=π∴A+(C/2)=π-B-(C/2)
∴sin[A+(C/2)]=4/5
∴A+(C/2)≠π/2
∴cos[A+(C/2)]cos[B+(C/2)]=-9/25【这步怎么来的?】
∴原式=(-9/25)+16/25=7/25
求回答.
在△ABC中,已知sin[B+(C/2)]=4/5,求cos(A-B)的值.【过程中有一步不懂,cos(A-B)=cos{[A+(C/2)]-[B+(C/2)]}=cos[A+(C/2)]cos[B+(C/2)]+sin[A+(C/2)]sin[B+(C/2)]∵sin[B+(C/2)]=4/5∵A+B+C=π∴A+(C/2)=π-B-(C/2)∴sin[A+(C/2)]=4/5∴A+(C/2)≠
设cos[A+(C/2)]=a a²=9/25
A+(C/2)=π-B-(C/2) cos[A+(C/2)]=-cos[B+(C/2)]
cos[A+(C/2)]cos[B+(C/2)]=-a²=-9/25
sin[B+(C/2)]=4/5,B+C/2 ≈ 53° 或者 127°
sin[A+(C/2)]=4/5,A+C/2 ≈ 53° 或者 127°
则 A+(C/2) 和B+(C/2)必有一个为钝角、一个为锐角且互补
因为A+(C/2)+B+(C/2) = A+B+C=180°
即cos[B+(C/2)]和cos[A+(C/2)] 为一正一负,即os[B+(C/2)]×cos[A+(C/2)] = - 3/5 * 3/5= -9/25