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 yтАж..(i)
So,
sin 5x = sin 3x cos 2x + cos 3x sin 2x
= sin (2x + x) cos 2x + cos (2x + x) sin 2xтАжтАж..(ii)
And
cos (x + y) = cos x cos y тАУ sin x sin yтАжтАж(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 тАУ sin 2x sin x) sin 2x тАж (iv)
Now sin 2x = 2sin x cos xтАжтАжтАж(v)
And cos 2x = cos2x тАУ sin2xтАжтАжтАж(vi)
Substituting equation (v) and (vi) in equation (iv),we get
sin 5x = [(2 sin x cos x) cos x + (cos2x тАУsin2x) sin x] (cos2x тАУ sin2x) + [(cos2xтАУ sin2x) cos x тАУ (2 sin x cos x) sin x)] (2 sin x cos x)
= [2 sin x cos2┬аx + sin x cos2xтАУ sin3x] (cos2x тАУ sin2x) + [cos3x тАУsin2x cos x тАУ 2 sin2┬аx cos x] (2 sin x cos x)
= cos2x [3 sin x cos2┬аx тАУsin3x] тАУ sin2x [3 sin x cos2┬аx тАУ sin3x]+ 2 sin x cos4x тАУ 2 sin3┬аx cos2┬аx тАУ4 sin3┬аx cos2┬аx
= 3 sin x cos4┬аx тАУ sin3xcos2x тАУ 3 sin3┬аx cos2┬аx тАУ sin5x+ 2 sin x cos4x тАУ 2 sin3┬аx cos2┬аx тАУ4 sin3┬аx cos2┬аx
= 5 sin x cos4┬аx тАУ10sin3xcos2x+sin5x
= RHS
Hence proved.