作业帮已知f(x)=sin(π-x)sin(π/2-x)+cos²x
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/12 05:40:31
已知f(x)=sin(π-x)sin(π/2-x)+cos²x
当-π/8≤x≤3π/8,求f(x)的单调区间,
当-π/8≤x≤3π/8,求f(x)的单调区间,
f(x)=sin(π-x)sin[(π/2)-x]+cos²x
=sinxcosx+cos²x
=(1/2)sin2x+[(cos2x+1)/2]
=(1/2)(sin2x+cos2x)+(1/2)
=(1/2)*(√2)*sin[2x+(π/4)]+(1/2)
=(√2/2)sin[2x+(π/4)]+(1/2)
当2x+(π/4)∈[2kπ-(π/2),2kπ+(π/2)]时,f(x)单调递增
即,x∈[kπ-(3π/8),kπ+(π/8)]
当2x+(π/4)∈[2kπ+(π/2),2kπ+(3π/2)]时,f(x)单调递增
即,x∈[kπ+(π/8),kπ+(5π/8)]
而,x∈[-3π/8,3π/8]
所以：
x∈[-π/8,π/8]时,f(x)单调递增；
x∈[π/8,3π/8]时,f(x)单调递减.
=sinxcosx+cos²x
=(1/2)sin2x+[(cos2x+1)/2]
=(1/2)(sin2x+cos2x)+(1/2)
=(1/2)*(√2)*sin[2x+(π/4)]+(1/2)
=(√2/2)sin[2x+(π/4)]+(1/2)
当2x+(π/4)∈[2kπ-(π/2),2kπ+(π/2)]时,f(x)单调递增
即,x∈[kπ-(3π/8),kπ+(π/8)]
当2x+(π/4)∈[2kπ+(π/2),2kπ+(3π/2)]时,f(x)单调递增
即,x∈[kπ+(π/8),kπ+(5π/8)]
而,x∈[-3π/8,3π/8]
所以：
x∈[-π/8,π/8]时,f(x)单调递增；
x∈[π/8,3π/8]时,f(x)单调递减.
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