sin2x+2sin(45°+x)的最小值

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sin2x+2sin(45°+x)的最小值
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sin2x+2sin(45°+x)的最小值
sin2x+2sin(45°+x)的最小值

sin2x+2sin(45°+x)的最小值
sin2X + 2sin(X + 45°)
=2sinXcosX + 2(√2/2 sinX + √2/2 cosX)
=2sinXcosX + √2(sinX + cosX)
=2sinXcosX + √2(sinX + cosX) +1 - 1
=2sinXcosX + √2(sinX + cosX) + sin^2X + cos^2X - 1
=(sinX + cosX)^2 + √2(sinX + cosX) - 1
设(sinX + cosX) = u
原式= u^2 + √2 u - 1
= (u +√2/2)^2 - 1 - (√2/2)^2
= (u +√2/2)^2 - 3/2 ≥ -3/2
注意要检验u能否取值取到 令(u +√2/2)=0 即u= -√2/2
经检验可以 则 sin2X + 2sin(45+X)最小值为-3/2

sin2x+2sin(45°+x)
=2sinxcosx+√2[sinx+cosx]
=(sinx+cosx)^2+√2[sinx+cosx]-1
设 y=sinx+cosx
则上式可以化为
=y^2+√2y-1
=(y+√2/2)^2-3/2
sinx+cosx=√2sin(45°+x)
所以 -√2≤y≤√2
所以上式的最小值 是当 y=-√2/2 时取得 为 -3/2