高中数学题若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是( )A.cos(x+π/4) B.-cos(x-π/4) C.-cos(x+π
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![高中数学题若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是( )A.cos(x+π/4) B.-cos(x-π/4) C.-cos(x+π](/uploads/image/z/1697983-7-3.jpg?t=%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E9%A2%98%E8%8B%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%92%8Cy%3Dsin%28x%2B%CF%80%2F4%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%85%B3%E4%BA%8E%E7%82%B9P%28%CF%80%2F4%2C0%29%E5%AF%B9%E7%A7%B0%2C%E5%88%99f%28x%29%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E6%98%AF%E8%8B%A5%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%92%8Cy%3Dsin%28x%2B%CF%80%2F4%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%85%B3%E4%BA%8E%E7%82%B9P%28%CF%80%2F4%2C0%29%E5%AF%B9%E7%A7%B0%2C%E5%88%99f%28x%29%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F%E6%98%AF%28+%29A.cos%28x%2B%CF%80%2F4%29+B.-cos%28x-%CF%80%2F4%29+C.-cos%28x%2B%CF%80)
高中数学题若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是( )A.cos(x+π/4) B.-cos(x-π/4) C.-cos(x+π
高中数学题若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是
若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是( )
A.cos(x+π/4) B.-cos(x-π/4) C.-cos(x+π/4) D.cos(x-π/4)
高中数学题若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,则f(x)的表达式是( )A.cos(x+π/4) B.-cos(x-π/4) C.-cos(x+π
若函数y=f(x)的图象和y=g(x)的图象关于点P(a,b)对称,
f(a+x)+g(a-x)=2b
f(x)+g(2a-x)=2b;f(x)=2b-g(2a-x)
若函数y=f(x)的图象和y=sin(x+π/4)的图象关于点P(π/4,0)对称,
则f(x)=0-sin(π/2-x+π/4)=-cos(x-π/4)
c,先把原函数的图划出,再把图绕那个点转180度,得到所求的函数的图象,再代入一个周期内的五个点检验,就可以找出c是符合的…
设y=f(x)的图象上点(x,y),关于(π/4,0)的对称点为(x0,y0)
则:(x+x0)/2=π/4, (y+y0)/2=0
x0=π/2-x, y0=-y
因:y0=sin(x0+π/4)
所以 -y=sin(π/2-x+π/4)=cos(x-π/4)
y=-cos(x-π/4)