将Y=4cos^4x+4sin^4x-3化为y=Asin(wx+@)的形式,其结果为?,当Y＜0时,X的取值范围是?
问题描述：
将Y=4cos^4x+4sin^4x-3化为y=Asin(wx+@)的形式,其结果为?,当Y＜0时,X的取值范围是?
题目中＠为一个希腊字母,恕愚人才疏学浅,不会打出来．．．．拜托了!
答
Y=4cos^4x+4sin^4x-3
=4cos^4x+4sin^4x+8cos^2x*sin^2x-8cos^2x*sin^2x-3
=4(cos^2x+sin^x)^2-8cos^2x*sin^2x-3
=1-8cos^2x*sin^2x
=1-2(sin2x)^2
=cos4x
=sin(-4x+π/2)
Y＜0就是
cos4x2kπ+π/2kπ/2+π/8