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求函数y=sin^8(x)+cos^8(x)的最小值
人气:359 ℃ 时间:2019-08-17 10:25:47
解答
y=sin^8(x)+cos^8(x)
=[sin^4(x)]^2+[cos^4(x)]^2
=[sin^4(x)+cos^4(x)]^2-2sin^4(x)cos^4(x)
={[sin^2(x)+cos^2(x)]^2-2sin^2(x)cos^2(x)}^2
-(1/8)[sin(2x)]^4
=[1-(1/2)(sin2x)^2]^2-(1/8)[(1-cos4x)/2]^2
=[1-(1/2)(1-cos4x)/2]^2-(1/32)(1-2cos4x+cos4²x)
=9/16+(3/8)cos4x+(1/16)cos²4x-1/32+(1/16)cos4x-(1/32)cos²4x
=(3/32)cos²4x+(7/16)cos4x+17/32
=[3cos²4x+14cos4x+17]/32
y=(3t²+14t+17)/32,|t|
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