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已知sin^4A+Cos^4A=5/9,求sin2A的值

sin^4a+cos^4a =(sin²a+cos²a)²-2sin²acos²a =1-2(sinacosa)²=5/9 (sinacosa)²=2/9 第三象限 sina

因为sin2a=3/5=2sinacosa,sin²a+cos²a=1 所以sin^4a+cos^4a=(sin²a+cos²a)²-2sin²acos²a =1²-sin2a/2 =1-9/50 =41/50

cos2A=3/5 (sin2A)^2 = 1- 9/25 = 16/25 (sinA)^4+(cosA)^2 =[(sinA)^2+(cosA)^2]^2 - 2(sinAcosA)^2 =1- (1/2)(sin2A)^2 =1- (1/2)(16/25) =17/25

请查一下题目是否正确? [1-(sinA)^2-(cosA)^4]/[2(sinA)^2.(cosA)^2 ] =[ (1-(cosA)^2).(1+(cosA)^2) - (sinA)^2 ]/[2(sinA)^2.(cosA)^2 ] =[ (sinA)^2.(1+(cosA)^2) - (sinA)^2 ]/[2(sinA)^2.(cosA)^2 ] =[ (sinA)^2.(1+(cosA)^2-1) ]/[2(sinA...

解sin^4 A+cos^4 A =(sin^2 A-cos^2 A)^2+2sin^2Acos^2A =(-(cos^2 A-sin^2 A))^2+1/2×2sinAcosA×2sinAcosA =(-cos2A)^2+1/2sin2A×2sin2A =(cos2A)^2+1/2(sin2A)^2 =(cos2A)^2+1/2[1-(cos2A)^2] =(3/5)^2+1/2[1-(3/5)^2] =9/25+1/2×16/25 =9/...

2sin²acos²a =1/2x4sin²acos²a =1/2x(2sinacosa)² =1/2x(sin2a)² 二倍角公式 =1/2xsin²2a

解:(1) 当 sin a =0 时, cos a = ±√ [ 1 -(sin a)^2 ] = ±1, 满足 (sin a)^4 +(cos a)^4 =1. 此时 sin a +cos a = ±1. (2) 当 sin a = ±1 时, cos a = ±√ [ 1 -(sin a)^2 ] =0. 满足 (sin a)^4 +(cos a)^4 =1. 此时 sin a +cos a = ±1. (3) 当 ...

sin^4A+cos^4A=(+cos^2A)-2sin^2Acos^2A=1-1/2 sin^2A=1-1/2(1/4)^2=31/32

sina-cosa=1/5 等号两边同平方得 sin²a-2sinacosa+cos²a=1/25 即1-sin2a=1/25,∴sin2a=24/25>0 ∴0

sin^4a+cos^4a=(sin²a+cos²a)²-2sin²acos²a=1 1-2sin²acos²a=1 sin²acos²a=0 所以sina=0或cosa=0 sina=0,则cosa=±1 cosa=0,则sina=±1 所以 sina-cosa=1或-1

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