三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

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三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x
三角等式
求证:
cos^6x+sin^6x=1-3sin^2x+3sin^4x

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x
用公式a³+b³=(a+b) (a²-ab+b²)
cos^6x+sin^6x
=(cos²x)³ + (sin²x)³
=(cos²x+sin²x)[(cos²x)² - cos²x * sin²x + (sin²x)²]
=(1-sin²x)² - (1-sin²x) * sin²x + sin^4x
=1-2sin²x + sin^4x -sin²x + sin^4x + sin^4x
=1-3sin²x +3sin^4x

等号左边:cos^6x+sin^6x=(sin^2x+cos^2x)^3-3sin^4xcos^2x-3sin^2cos^4x=1-3sin^2xcos^2x(sin^2x+cos^2x)
=1-3sin^2cos^2x
等号右边:1-3sin^2x+3sin^4x=1-3sin^2x(1-sin^2x)=1-3sin^2xcos^2
所以左边=右边,得证。