Question

急求三角函数问题

已知函数f(x)=(cosx)^4-2sinxcosx-(sinx)^4.
1.求函数f(x)的最小正周期；
2.若x∈[0,π/2],求f(x)的最大值和最小值.

Answer 1

f(x)=(cosx)^4-2sinxcosx-(sinx)^4=(cos^2x)^2-(sin^2x)^2-sin2x-=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x=cos2x-sin2x=√2(√2/2cos2x-√2/2sin2x)=√2cos(2x+π/4)
1.函数f(x)的最小正周期T=2π/2=π
2.因为x∈[0,π/2]
所以2x∈[0,π]
2x+π/4∈[π/4，5π/4]
cos(2x+π/4)∈[-1,√2/2]
2cos(2x+π/4)∈[-2,√2]