高中数学题第二步怎么做
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f(x) = sin(x+phi)cos(phi)+cos(x+phi)sin(phi) - 2sin(phi)cos(x+phi)
=sin(x+phi)cos(phi)-cos(x+phi)sin(phi) = sin( x+phi - phi) = sin x
y = f(wx+π/4) = sin( wx + π/4)
在 2kπ+π/2<wx + π/4<2kπ+3π/2 是减函数(k为整数)
或:递减区间为 2kπ+π/4<wx <2kπ+5π/4
即 2kπ+π/4<=w π/2 < w π <=2kπ+5π/4
4k+1/2<=w<=2k+5/4
考虑 k 是整数,w>0,所以所求范围为: 1/2 <=w<=5/4
=sin(x+phi)cos(phi)-cos(x+phi)sin(phi) = sin( x+phi - phi) = sin x
y = f(wx+π/4) = sin( wx + π/4)
在 2kπ+π/2<wx + π/4<2kπ+3π/2 是减函数(k为整数)
或:递减区间为 2kπ+π/4<wx <2kπ+5π/4
即 2kπ+π/4<=w π/2 < w π <=2kπ+5π/4
4k+1/2<=w<=2k+5/4
考虑 k 是整数,w>0,所以所求范围为: 1/2 <=w<=5/4
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