> 数学 >
函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.请写出过程,
人气:121 ℃ 时间:2020-04-01 08:54:33
解答
y=cos(2x-3π/4)-2√2(sinx)^2
=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2
=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2
=√2/2*cos2x+√2/2*sin2x-√2
=cos(2x-pi/4)-√2
T=2pi/2=pi
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版