求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2

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求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2求证:(sinα)^4+(cosα)^4=1-1\2*(

求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2
求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2

求证:(sinα)^4+(cosα)^4=1-1\2*(sin2α)^2
因为sin²a+cos²a=1
所以(sin²a+cos²a)²=1=(sina)^4+(cosa)^4+2sin²a*cos²a
因为sin2a=2sinacosa
所以1/2(sin2a)²=2sin²a*cos²a
所以1-1/2sin²2a=(sina)^4+(cosa)^4+2sin²a*cos²a-2sin²a*cos²a=(sina)^4+(cosa)^4
等式 左边=右边
等式成立,得证

(sinα)^4+(cosα)^4
=[(sinα)^2+(cosα)^2]^2-2(sinα)^2(cosα)^2
=1-1/2*(2sinαcosα)^2
=1-1/2*(sin2α)^2