4 4sinxcosx 2sinx 2cosx的最大值是多少?
4+4sinxcosx+2sinx+2cosx=2(sinx+cosx)^2+2(sinx+cosx)+2
=2(sinx+cosx+1/2)^2+3/2
=2{(√2)[sinxcos(π/4)+cosxsin(π/4)]+1/2}^2+3/2
=2[√2sin(x+π/4)+1/2]^2+3/2
因为sin(x+π/4)≤1,
所以2[√2sin(x+π/4)+1/2]^2+3/2≤2(1/2+√2)^2+3/2,
即4+4sinxcosx+2sinx+2cosx的最大值=2(1/2+√2)^2+3/2=6+2√2。
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4+4sinxcosx+2sinx+2cosx=2(sinx+cosx)^2+2(sinx+cosx)+2
=2(sinx+cosx+1/2)^2+3/2
=2{(√2)[sinxcos(π/4)+cosxsin(π/4)]+1/2}^2+3/2
=2[√2sin(x+π/4)+1/2]^2+3/2
因为sin(x+π/4)≤1,
所以2[√2sin(x+π/4)+1/2]^2+3/2≤2(1/2+√2)^2+3/2,
即4+4sinxcosx+2sinx+2cosx的最大值=2(1/2+√2)^2+3/2=6+2√2。
。收起