maximum value of sin^6 +cos^6
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(sin²)³+(cos²)³
= (sin²+cos²)(sin⁴-sin²cos²+cos²)
= sin⁴+cos⁴-sin²cos²
=(sin²)²+(cos²)²-sin²cos²[a²+b²=(a+b)²-2ab]
= (sin²+cos²)² -2sin²cos²-sin²cos²
= (1)²-3sin²cos² = 1 - 3sin²cos²
= (sin²+cos²)(sin⁴-sin²cos²+cos²)
= sin⁴+cos⁴-sin²cos²
=(sin²)²+(cos²)²-sin²cos²[a²+b²=(a+b)²-2ab]
= (sin²+cos²)² -2sin²cos²-sin²cos²
= (1)²-3sin²cos² = 1 - 3sin²cos²
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