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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

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

已知sin(π+a)=4/5 那么sina=-4/5 又a是第四象限角 那么cosa=√[1-(-4/5)^2]=3/5 所以cos(a-2π)=cosa=3/5 如果不懂,请追问,祝学习愉快!

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

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

(sina)^2=1-(cosa)^2,所以(sina)^4=[1-(cosa)^2]^2=(cosa)^4-2(cosa)^2+1, 所以 sin的四方a-sin平方a+cos平方a=(cosa)^4-2(cosa)^2+1-[1-(cosa)^2]+(cosa)^2=(cosa)^4

sin^4a+cos^4a =(sin²a+cos²a)²-2sin²acos²a =1-(1/2)(2sinacosa)² =1-(1/2)sin²2a=5/9 sin²2a=8/9 a为第三象限角 2kπ+π

由已知得: sin^4a-cos^4a=(sin²a+cos²a)(sin²a-cos²a)=-cos2a=-3/5 则有:cos2a=3/5 即得:2cos²a-1=3/5 cos²a=4/5 已知3π/2

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

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