Question

高一 数学 高中数学题 请详细解答,谢谢! (31 6:54:14)

求函数y=sin(π/3+4x)+cos(4x-π/6)的单调区间

Answer 1

y=sin(π/3+4x)+cos(4x-π/6)
=sin(π/3+4x)+sin(π/2-4x+π/6)
=sin(π/3+4x)+sin(2π/3-4x)
=2sin[(π/3+4x+2π/3-4x)/2]cos[(π/3+4x-2π/3+4x)/2]
=2sin(π/2)cos(4x-π/6)
=2cos(4x-π/6)

当2kπ<=4x-π/6<=2kπ+π时，
即kπ/2+π/24<=x<=kπ/2+7π/24时（k为整数）单调递减

当（2k-1）π<=4x-π/6<=2kπ时，
即kπ/2-5π/24<=x<=kπ/2+π/24时（k为整数），单调递增

Answer 2

把这个函数的两项按和、差公式展开，再合并同类项后结合成一个函数，单调区间立得。

Answer 3

先化简,再求值.具体如下:
y=sin(π/3+4x)+cos(4x-π/6)
=sinπ/3cos4x+cosπ/3sin4x+cos4xcosπ/6+sin4xsinπ/6
=√3*cos4x+sin4x
=2sin(4x+π/3)
由-π/2+2kπ<=4x+π/6<=π/2+2kπ=====>-π/6+kπ/2<=x<=π/12+kπ/2为单调增区间
由π/2+2kπ<=4x+π/6<=3π/2+2kπ=====>π/12+kπ/2<=x<=π/3+kπ/2为单调增区间

Answer 4

y=sin(π/3+4x)+cos(4x-π/6)
=sinπ/3cos4x+cosπ/3sin4x+cos4xcosπ/6+sin4xsinπ/6
=√3*cos4x+sin4x
=2sin(4x+π/3)
增区间：-π/2+2kπ<=4x+π/3<=π/2+2kπ 【k为整数】
即kπ/2-5π/24<=x<=kπ/2+π/24
减区间：π/2+2kπ<=4x+π/3<=3π/2+2kπ
即kπ/2+π/24<=x<=kπ/2+7π/24