Question -
Answer -
Let us consider LHS:
sin 5x
Now,
sin 5x = sin (3x + 2x)
But we know,
Sin (x + y) = sin x cos y + cos x sin y…..(i)
So,
sin 5x = sin 3x cos 2x + cos 3x sin 2x
= 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 + ( cos2x 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)
= [2 sin x cos2 x + sin x cos2x– sin3x] (cos2x – sin2x) + [cos3x –sin2x cos x – 2 sin2 x cos x] (2 sin x cos x)
= 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
= 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
= 5 sin x cos4 x –10sin3xcos2x+sin5x
= RHS
Hence proved.