Sin8A/SinA=2cosAcos2Acos8A
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generally the formula for sin 2x = 2sinx•cosx => sin8A = sin 2(4A) = 2 sin4A cos4A =2sin2(2A) cos4A=2sin2A cos2A cos4A
=>2sinA cosA cos2A cos4A----(1)
dividing (1) with sinA we get ,,
sin8A /sinA = 2 cosA cos2A cos4A. according to my knowledge question might be wrong as cos 8A it might be cos4A instead cos8A
=>2sinA cosA cos2A cos4A----(1)
dividing (1) with sinA we get ,,
sin8A /sinA = 2 cosA cos2A cos4A. according to my knowledge question might be wrong as cos 8A it might be cos4A instead cos8A
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