Answer :

LHS is


sin 5x = sin(3x+2x)


But we know,


sin(x+y) = sin x cos y+cos x sin y…..(i)


sin 5x = sin 3x cos 2x+cos 3x sin 2x


sin 5x = sin (2x+x) cos 2x+cos (2x+x) sin 2x……..(ii)


And


cos (x+y) = cos(x)cos(y) – sin(x)sin(y)……(iii)


Now substituting equation (i) and (iii) in equation (ii), we get


sin 5x = (sin 2x cos x+cos 2x sin x )(cos 2x)+( cos 2x cos x – sin 2x sin x) (sin 2x)……..(iv)


Now sin 2x = 2sin x cos x………(v)


And cos 2x = cos2x –sin2x………(vi)


Substituting equation (v) and (vi) in equation (iv), we get


sin 5x =[(2 sin x cos x)cos x+(cos2x–sin2x)sin x]( cos2x–sin2x)+[( cos2x–sin2x)cos x – (2 sin x cos x) sin x)]( 2 sin x cos x)


sin 5x =[2 sin x cos2 x+sin xcos2x–sin3x]( cos2x–sin2x)+[cos3x–sin2xcos x – 2 sin2 x cos x]( 2 sin x cos x)


sin 5x =cos2x[3 sin x cos2 x –sin3x]–sin2x[3 sin x cos2 x–sin3x]+2 sin x cos4x–2 sin3 x cos2 x – 4 sin3 x cos2 x


sin 5x = 3 sin x cos4 x –sin3xcos2x– 3 sin3 x cos2 x–sin5x +2 sin x cos4x–2 sin3 x cos2 x – 4 sin3 x cos2 x


sin 5x = 5 sin x cos4 x –10sin3xcos2x +sin5x


Hence LHS = RHS


[Hence proved]


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