求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/19 11:04:32
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期

求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期
求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期

求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期
这里需要公式:cos(π-x)=sinx,sin(π/2-x)=cosx
y=[sin2x+sin(2x+π/3)-cos(π/2-2x﹚]/[cos2x+cos﹙2x+π/3﹚-sin(π/2-2x)]
=[sin2x+sin(2x+π/3)-sin2x]/[cos2x+cos﹙2x+π/3﹚-cos2x]
=sin(2x+π/3)/cos﹙2x+π/3﹚
=tan(2x+π/3)
即y=tan(2x-π/3)
所以最小正周期为:π/2.

y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]
={sin2x+sin【2x+(π/3)】-cos【(π/2)-2x】}/{cos2x+cos【2x+π/3】-sin【(π/2)-2x】}
={sin2x+sin【2x+(π/3)】-sin2x}/{cos2x+c...

全部展开

y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]
={sin2x+sin【2x+(π/3)】-cos【(π/2)-2x】}/{cos2x+cos【2x+π/3】-sin【(π/2)-2x】}
={sin2x+sin【2x+(π/3)】-sin2x}/{cos2x+cos【2x+(π/3﹚】-cos2x}
=sin【2x+(π/3)】/cos【2x+(π/3)】
=tan【2x+(π/3)】
∴最小正周期T=2π/2=π

收起

求函数y=[sin2x+sin(2x+π/3)-cos(1/2)(π-4x)]/[cos2x+cos(2x+π/3﹚-sin(1/2)(π-4x)]的最小正周期
y=[sin2x+sin(2x+π/3)-cos(π/2-2x)]/[cos2x+cos(2x+π/3)-sin(π/2-2x)]
=[sin2x+sin(2x+π/3)-sin2x]/[cos2x+cos(2x+π/3)-cos2x]
=sin(2x+π/3)/cos(2x+π/3)
=tan(2x+π/3)
故其最小正周期为π/2.

把题目中的复杂的单项分解成简单的单项,然后合并同类项。最后就结合成基本三角函数的形式。sin(x),cos(x),等等 。