ula_p
幼苗
共回答了15个问题采纳率:93.3% 举报
你这里是不是少打个条件x∈[0,π/2]啊
f(x)=cos^4x-2sinxcosx-sin^4x
=(cosx)^2-(sinx)^2-2sinxcosx
=cos2x-sin2x
=(√2)cos(2x+π/4),
(1)f(x)的最小正周期是π.
(2)x∈[0,π/2],
∴2x+π/4∈[π/4,5π/4],
∴f(x)的最小值是-√2,这时2x+π/4=π,x=3π/8.
1年前
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ula_p
f(x)的最大值就是当2X+π/4=1 即2Kπ+π/2=2X+π/4 2Kπ+π/2=2X+π/4 2Kπ+π/2-π/4=2X 2X=2Kπ+π/4 X=Kπ+8/π(K∈Z) 此时f(x)max=√2 f(x)的最小值就是当2X+π/4=-1 即2Kπ+π=2X+π/4 2Kπ+π/2-π/4=2X 2X=2Kπ+π/4 X=Kπ+π/8(K∈Z) 此时f(x)min=-√2 不懂的欢迎追问,如有帮助请采纳,谢谢!