If the expression 4 sin5x * cos3x * cos2x is expressed as the sum of three sines then two of them are sin 4x and sin 10x. The third one is ...
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LHS = 4 Sin 5x * Cos 3x * Cos 2x
We know 2 Sin A Cos B = Sin (A+B) + Sin (A-B)
2 Sin 5x Cos 2x = Sin 7x + Sin 3x
LHS = 2 * Cos 3x [ sin 7x + Sin 3x ]
= 2 Sin 7 x Cos 3 x + 2 Sin 3 x Cos 3x
= Sin 10 x + Sin 4 x + Sin 6x
as 2 Sin A Cos A = Sin 2A
Hence, the answer is Sin 6x
We know 2 Sin A Cos B = Sin (A+B) + Sin (A-B)
2 Sin 5x Cos 2x = Sin 7x + Sin 3x
LHS = 2 * Cos 3x [ sin 7x + Sin 3x ]
= 2 Sin 7 x Cos 3 x + 2 Sin 3 x Cos 3x
= Sin 10 x + Sin 4 x + Sin 6x
as 2 Sin A Cos A = Sin 2A
Hence, the answer is Sin 6x
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