在三角形ABC中,cos^2A+cos^2B+cos^2C=1,则三角形的形状是?cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)=2(sinBcosCcosBsinC) cos^2Bcos^2C+cos^2Ccos^2B=2(sinBcosCcosBsinC) 这两步是如何变化的?请用过程具体说明!
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 19:29:05
![在三角形ABC中,cos^2A+cos^2B+cos^2C=1,则三角形的形状是?cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)=2(sinBcosCcosBsinC) cos^2Bcos^2C+cos^2Ccos^2B=2(sinBcosCcosBsinC) 这两步是如何变化的?请用过程具体说明!](/uploads/image/z/1600503-15-3.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ccos%5E2A%2Bcos%5E2B%2Bcos%5E2C%3D1%2C%E5%88%99%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%BD%A2%E7%8A%B6%E6%98%AF%3Fcos%5E2B%2Bcos%5E2C-%28sin%5E2Bcos%5E2C%2Bcos%5E2Bsin%5E2C%29%3D2%28sinBcosCcosBsinC%29+cos%5E2Bcos%5E2C%2Bcos%5E2Ccos%5E2B%3D2%28sinBcosCcosBsinC%29+%E8%BF%99%E4%B8%A4%E6%AD%A5%E6%98%AF%E5%A6%82%E4%BD%95%E5%8F%98%E5%8C%96%E7%9A%84%3F%E8%AF%B7%E7%94%A8%E8%BF%87%E7%A8%8B%E5%85%B7%E4%BD%93%E8%AF%B4%E6%98%8E%21)
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