> 数学 >
函数f(x)=cos(2x-π/3)+cos(2x+π/6)的最大值和最小正周期以及化简过程
人气:478 ℃ 时间:2019-11-23 07:05:02
解答
f(x)=cos(2x-π/3)+cos(2x+π/6)
=cos(2x-π/3)+sin(-2x+π/3)
=cos(2x-π/3)-sin(2x-π/3)
=√ 2cos(2x-π/3+π/4)
=√ 2cos(2x-π/12)
所以函数f(x)的最大值为 √2,最小正周期为π为什么cos(2x+π/6)=sin(-2x+π/3)公式sina=sin(π/2-a)cos(2x+π/6)=sin(π/2-2x-π/6)=sin(-2x+π/3)
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版