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(00 +sin®) + 3sin® _ |
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Answer» sin^6x+cos^6x+3sin^2xcos^2x =sin^6x+3sin^2xcos^2x(1)+cos^6x =sin^6x+3sin^2xcos^2x(sin^2x+cos^2x)+cos^6x =sin^6x+3sin^4xcos^2x+3sin^2xcos^4x+cos^6x=(sin^2x+cos^2x)^3=1^3=1 |
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