作业帮已知f(x)=sin(π-x)sin(π/2-x)+cos²x当-π/8≤x≤3π/8,求f(x)的单调区间,
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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,
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,
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,
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.
f(x)=sin(π-x)sin(π/2-x)+(cosx)^2
= sinx.cosx + (cosx)^2
=(1/2)sin(2x) + (1/2)[ 1+cos(2x)]
=(√2/2)sin(2x+π/4) + 1/2
单调区间
增加=[-π/8, π/8 ]
减小= [π/8,3π/8]
答：
f(x)=sin(π-x)sin(π/2-x)+cos²x
=sinxcosx+(1/2)(cos2x+1)
=(1/2)sin(2x)+(1/2)cos(2x)+1/2
=(√2/2)*[(√2/2)sin(2x)+(√2/2)cos(2x)]+1/2
=(√2/2)sin(2x+π/4)+1/2
因为：-π/8<=x<=3π/8<...
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答：
f(x)=sin(π-x)sin(π/2-x)+cos²x
=sinxcosx+(1/2)(cos2x+1)
=(1/2)sin(2x)+(1/2)cos(2x)+1/2
=(√2/2)*[(√2/2)sin(2x)+(√2/2)cos(2x)]+1/2
=(√2/2)sin(2x+π/4)+1/2
因为：-π/8<=x<=3π/8
所以：-π/4<=2x<=3π/4
所以：0<=2x+π/4<=π
所以：单调递增区间满足0<=2x+π/4<=π/2
所以：-π/8<=x<=π/8
所以：单调递增区间为[-π/8，π/8]，单调递减区间[π/8，3π/8]
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f(x)=sin(π-x)sin(π/2-x)+cos²x
=sinxcosx + cos²x
=1/2(2sinxcosx) + 1/2( 2cos²x - 1) + 1/2
= sin2x/2 + cos2x/2 + 1/2
=√2/2 (√2/2 sin2x + √2/2 cos2x) + ...
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f(x)=sin(π-x)sin(π/2-x)+cos²x
=sinxcosx + cos²x
=1/2(2sinxcosx) + 1/2( 2cos²x - 1) + 1/2
= sin2x/2 + cos2x/2 + 1/2
=√2/2 (√2/2 sin2x + √2/2 cos2x) + 1/2
=√2/2 sin(2x + π/4) + 1/2
∵-π/8≤x≤3π/8
∴0≤2x + π/4≤π
故0≤2x + π/4≤π/2,
f(x)单调递增，此时 -π/8≤x≤π/8 区间：[-π/8,π/8]
π/2<2x + π/4≤π,
f(x)单调递减，此时 π/8
收起
f(x)=sinxcosx+cos²x
=(1/2)sin2x+(1+cos2x)/2
=(√2/2)sin(2x+π/4)+1/2