1.

\operatorname { cos } x + \operatorname { cos } 3 x + \operatorname { cos } 5 x + \operatorname { cos } 7 x = 4 \operatorname { cos } x \operatorname { cos } 2 x

Answer»

LHS=cos x+cos 3x+cos 5x+cos 7x=cos x + cos(7x) + cos(3x) + cos(5x)=cos(4x - 3x) + cos(4x + 3x) + cos(4x - x) + cos(4x + x)= 2 cos(4x)cos(3x) + 2cos(4x)cos(x) (Using the formula 1)= 2 cos(4x) [cos(3x) + cosx]= 2 cos(4x)[cos(2x + x) + cos(2x - x)]= 2cos(4x)[2cos(2x)cosx] (Using the formula 1)=4 cos(4x) cos(2x) cos(x) = 4cosxcos2xcos4x = RHS


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