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已知sin^4A+Cos^4A=5/9,求sin2A的值

sin^4a+cos^4a=(sin^2a+cos^2a)^2-2sin^2acos^2a=5/9 1-2sin^2acos^2a=5/9 sin^2acos^2a=2/9 sin^2acos^2a=(sin2a)^2/4=2/9 (sin2a)^2=8/9 sin2a=±2根号2/3

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

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

tanα+cotα=sinα/cosα+cosα+sinα=(sin^2α+cos^2α)/cosαsinα=1/sinαcosα =2/sin2α sin^2α=(1-cos2α)/2 cos^2α=(1+cos2α)/2 sin^4α+cos^4α=[(1-cos2α)/2]^2+[(1+cos2α)/2]^2 =1/2+1/2*cos^2(2α)=5/9 cos^2(2α)=1/9 sin^2(2α)=1-cos^2(2α)=8/9 π

高二的题?

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

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/...

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...

cos2a=cos^2a-sin^2a=1-2sin^a=4/5 ,sin^2a=1/10, cos^a=9/10 (cos^2a-sin^2a)^2=16/25=cos^4a+sin^4a-2cos^2asin^2a cos^4a+sin^4a=16/25+2cos^2asin^2a=16/25+9/50=41/50

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

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