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已知sin^4A+Cos^4A=5/9,求sin2A的值

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

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 ...

证: sin⁴a+cos⁴a =(sin²a)²+(cos²a)²+2sin²acos²a-2sin²acos²a =(sin²a+cos²a)²-2sin²acos²a =1²-2sin²acos²a =1-2sin²acos²a 等式成...

原式=(cos^2a+sin^2a)(cos^2a-sin^2a) =1*(cos2a) =cos2a

(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)sin²2a =1-(1/2)[1-cos²2a] =1-(1/2)[1-(2/9)] =11/18 这样可以么?

倍角公式是sin2a=2sinacosa 由和角公式得sin2a=sinacosa+cosasina=2sinacosa

(1 + sin 4 a - cos 4 a) / (1 + sin 4 a + cos 4 a) = (1 - cos 4 a + sin 4 a) / (1 + cos 4 a + sin 4 a) = (2 sin^2 2 a + 2 sin 2 a cos 2 a)/ (2 cos^2 2 a + 2 sin 2 a cos 2 a) = [2 (sin 2 a) (sin 2 a + cos 2 a)]/ [2 (cos 2 a) (cos...

sin4a-cos4a=sin²a-cos²a, 这里,sin4a中的4是4次方,(sina)^4,不是a*4 这样写清晰一些: (sina)^4-(cosa)^4 =[(sina)^2+(cosa)^2]*[(sina)^2-(cosa)^2] (sina)^2+(cosa)^2=1 =(sina)^2-(cosa)^2 这样就理解了吧!

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