f（x）=2cos^2x+2sinxcosx+1.（1）x属于【派/24,7派/24】时,求f（x）的最值.（2）f（a

gracedeep2022-10-04 11:39:541条回答

f（x）=2cos^2x+2sinxcosx+1.（1）x属于【派/24,7派/24】时,求f（x）的最值.（2）f（a/2）=11/5,a属于（0,派）,求sin2a（要有过程喔,）

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beckhamshenhua 共回答了23个问题 | 采纳率87%: （1）f（x）=2cos²x+2sinxcosx+1
=2×(cos2x+1)/2+sin2x+1
=cos2x+sin2x+2
=√2(√2/2 cos2x+√2/2 sin2x)+2
=√2sin(2x+π/4）+2
∵x∈(π/24,7π/24)
∴ π/12 ﹤ 2x﹤7π/12
π/3 ﹤2x+π/4﹤5π/6
∴ 1/2 ﹤ sin(2x+π/4）﹤1
∴ √2 /2 +1﹤ f（x）﹤√2+2
(2)∵f（a/2）=11/5
∴f（a/2) =√2sin(a+π/4）+2=11/5
∴sin(a+π/4）=√2/10
√2/2 cosa+√2/2 sina=√2/10
sina+cosa=1/5
(sina+cosa)²=1/25
sin²a+2sinacosa+cos²a=1/25
1+sin2a=1/25
∴sin2a=-24/25; 1年前