sin^6+cos^6=1-3sin²×cos²
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formulas used :
a³+b³ = (a+b)(a²+b²-ab)
a²+b² = (a+b)²-2ab
sin²x + cos²x = 1
LHS
sin^6x + cos^6x
(sin²x)³ + (cos²x)³
=> (sin²x+cos²)(sin⁴x+cos⁴x-sin²xcos²x)
=> (1) ( (sin²x)² + (cos²)² - sin²xcos²x)
=> (sin²x+cos²x)²-2sin²xcos²x -sin²xcos²x
=> 1=3sin²xcos²x
therefore .LHS = RHS
hence proved
a³+b³ = (a+b)(a²+b²-ab)
a²+b² = (a+b)²-2ab
sin²x + cos²x = 1
LHS
sin^6x + cos^6x
(sin²x)³ + (cos²x)³
=> (sin²x+cos²)(sin⁴x+cos⁴x-sin²xcos²x)
=> (1) ( (sin²x)² + (cos²)² - sin²xcos²x)
=> (sin²x+cos²x)²-2sin²xcos²x -sin²xcos²x
=> 1=3sin²xcos²x
therefore .LHS = RHS
hence proved
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