f(x)=sin x的平方+sinxcosx-2sin(x+π/4)sin(x-π/4)若tanα=2求f(α...
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发布时间:2024-10-23 12:36
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热心网友
时间:4分钟前
f(x)=sin² x+sinxcosx-2sin(x+π/4)sin(-π/2+x+π/4)
= (1-cos2x)/2+1/2sin2x+2sin(x+π/4) cos(x+π/4)
= 1/2-1/2cos2x+1/2sin2x+sin(2x+π/2)
=1/2-1/2sin2x-1/2cos2x+cos2x
= 1/2+1/2sin2x+1/2cos2x
=1/2+√2/2sin(2x+π/4)
∵tanα=2 ∴sinα=2cosα
∵sin²α+cos²α=1 ∴cos²α=1/5
∴f(α)=1/2+1/2sin2α+1/2cos2α
=1/2+sinαcosα+1/2+cos²α
=1+2cos²α+cos²α=1+3/5=8/5
2
∵x∈[π/12,π/2]
∴2x∈[π/6,π]
∴2x+π/4∈[π/3,3π/4]
∴1/2≤√2/2sin(2x+π/4)≤√2/2
∴1≤f(x)≤(1+√2)/2
f(x)值域为[1,(1+√2)/2]
热心网友
时间:2分钟前
我算出来了。1.f(α)=3/5
2. F(x)∈ [0,(1+√2)\2]
你看答案对不对,对了的话,我可以跟你讲过程。放暑假,这做题都生疏了
1.
=(1-cos2x)/2+1/2sin2x+ cos2x
=1/2( sin2x+ cos2x) +1/2
因为tanα=2
所以sin2α=( 2sinαcosα) /(sin²α+cos²α)=4/5
cos2α=( cos²α- sin²α) /(sin²α+cos²α)= -3 /5
所以f(α)=1/2+1/2(sin2α+cos2α)= 3 /5
2. 可以由1得,f(x)=1/2+√2/2sin(2x+π/4)
由x∈[π/12,π/2]
得2x+π/4∈[5π/12,5π/4]
sin(2x+π/4) ∈[ -√2/2,1]
所以F(x)∈ [0,(1+√2)\2]
热心网友
时间:2分钟前
x=(sinx)^2+sinxcosx-2*[(sinxcos45+cosxsin45)*(sinxcos45-cosxsin45)]
=(sinx)^2+sinxcosx-[(sinx)^2-(cosx)^2]
=(cosx)^2+sinxcosx
=(1/2)[2(cosx)^2-1]+(1/2)*(2sinxcosx)+1/2
=cos2x+sin2x+1/2
=√2*sin(2x+π/4)+1/2
热心网友
时间:5分钟前