求函数y=sin²(x+π/12)+cos²(x-π/12)-1的最大值
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y=sin²(x+π/12)+cos²(x-π/12)-1
=(1-cos(2x+π/6))/2+(1+cos(2x-π/6))/2-1
=1/2[-cos(2x+π/6) +cos(2x-π/6) ]
=1/2[-cos2x cosπ/6+sin2x sinπ/6
+cos2x cosπ/6+sin2x sinπ/6]
= sin2x sinπ/6=1/2 sin2x≤1/2..
函数最大值是1/2.
=(1-cos(2x+π/6))/2+(1+cos(2x-π/6))/2-1
=1/2[-cos(2x+π/6) +cos(2x-π/6) ]
=1/2[-cos2x cosπ/6+sin2x sinπ/6
+cos2x cosπ/6+sin2x sinπ/6]
= sin2x sinπ/6=1/2 sin2x≤1/2..
函数最大值是1/2.
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