(sin 3x + sin x) sin x + (cos 3x - cos x) cos x = 0
prove it
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Answered by
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Step-by-step explanation:
L.H.S. = (sin 3x + sin x)sin x + (cos 3x - cos x)cos x
= sin 3x sin x + sin²x + cos 3x cos x - cos²x
= (cos 3x cos x + sin 3x sin x) - (cos²x -
sin²px)
= cos (3x – x) - pcos 2x
= cos 2x - cos 2x
= 0
= R.H.S.
Answered by
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formula: sinC+ sinD= 2sinC+D/2 × CosC-D/2
cosC- cosD= 2sinC+D/2× Sin D-C/2
the put this formula in your question
2sin2x cosx sinx+ 2sin2x sin(-x) cosx
2sin2x cosx sinx -2sin2x cosx sinx
( sin(-x)= -sinx)
=0= RHS
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