据任意角的三角函数的定义证明(sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)证明:左边=(y/r+y/x)•(x/r+x/y)=y(1/r+1/x)•x(1/r+1/y)=[y(1/y+1+r)]•[x(1/x+1/r)]=(1+y/r)(1+x/r)=右边
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![据任意角的三角函数的定义证明(sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)证明:左边=(y/r+y/x)•(x/r+x/y)=y(1/r+1/x)•x(1/r+1/y)=[y(1/y+1+r)]•[x(1/x+1/r)]=(1+y/r)(1+x/r)=右边](/uploads/image/z/5350564-28-4.jpg?t=%E6%8D%AE%E4%BB%BB%E6%84%8F%E8%A7%92%E7%9A%84%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E5%AE%9A%E4%B9%89%E8%AF%81%E6%98%8E%28sin%CE%B1%2Btan%CE%B1%29%28cos%CE%B1%2B1%2Ftan%CE%B1%29%3D%281%2Bsin%CE%B1%29%281%2Bcos%CE%B1%29%E8%AF%81%E6%98%8E%EF%BC%9A%E5%B7%A6%E8%BE%B9%3D%EF%BC%88y%2Fr%2By%2Fx%29%26%238226%3B%28x%2Fr%2Bx%2Fy%29%3Dy%281%2Fr%2B1%2Fx%29%26%238226%3Bx%281%2Fr%2B1%2Fy%29%3D%5By%281%2Fy%2B1%2Br%29%5D%26%238226%3B%5Bx%281%2Fx%2B1%2Fr%29%5D%3D%281%2By%2Fr%29%281%2Bx%2Fr%29%3D%E5%8F%B3%E8%BE%B9)
据任意角的三角函数的定义证明(sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)证明:左边=(y/r+y/x)•(x/r+x/y)=y(1/r+1/x)•x(1/r+1/y)=[y(1/y+1+r)]•[x(1/x+1/r)]=(1+y/r)(1+x/r)=右边
据任意角的三角函数的定义证明(sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)
证明:左边=(y/r+y/x)•(x/r+x/y)
=y(1/r+1/x)•x(1/r+1/y)
=[y(1/y+1+r)]•[x(1/x+1/r)]
=(1+y/r)(1+x/r)=右边
据任意角的三角函数的定义证明(sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)证明:左边=(y/r+y/x)•(x/r+x/y)=y(1/r+1/x)•x(1/r+1/y)=[y(1/y+1+r)]•[x(1/x+1/r)]=(1+y/r)(1+x/r)=右边
说明:
据任意角的三角函数的定义的意思是利用三角函数的定义
在直角坐标系中,三角函数的定义是
sinα=y/r,cosα=x/r,tanα=y/x
——(x,y)是α角终边上点的坐标,r是这一点到坐标原点的距离
把这几个式子带入置换化简便是,即
左边=(sinα+tanα)(cosα+1/tanα)
=(y/r+y/x)•(x/r+x/y)
=y(1/r+1/x)•x(1/r+1/y)
=[y(1/y+1/r)]•[x(1/x+1/r)] ——这一步,x、y、(1/r+1/y)、(1/r+1/x)四个因式重新组合了一下.
另外,你原来的式子里把1/r写成1+r了,我给改过来了.
=(1+y/r)(1+x/r) ——这一步是将x、y乘进括号里就行了
右边=(1+sinα)(1+cosα)
=(1+y/r)(1+x/r)
显然,左边=右边
即 (sinα+tanα)(cosα+1/tanα)=(1+sinα)(1+cosα)
证毕
(sinα+tanα)(cosα+1/tanα) 展开,并且tanα=sinα/cosα,得cosα*sinα+sinα+cosα+1
=(1+sinα)(1+cosα)
你坑定是懒的化简