As we know that,
cos 3θ = 4cos3θ – 3cosθ
Therefore, 4 cos3θ = cos3θ + 3cosθ
cos3 θ = [cos3θ + 3cosθ]/4 …… (i)
Now similarly,
sin 3θ = 3sin θ – 4sin3 θ
4 sin3θ = 3sinθ – sin 3θ
sin3θ = [3sinθ – sin 3θ]/4 …….. (ii)
Then,
Let us consider the LHS
cos3 x sin 3x + sin3 x cos 3x
By substituting the values from equation (i) and (ii), we get
cos3 x sin 3x + sin3 x cos 3x = (cos 3x + 3 cos x)/4 sin 3x + (3sin x – sin 3x)/4 cos 3x
= 1/4 (sin 3x cos 3x + 3 sin 3x cox x + 3sin x cos 3x – sin 3x cos 3x)
= 1/4 (3(sin 3x cos x + sin x cos 3x) + 0)
= 1/4 (3 sin (3x + x))
(As we know, sin(x + y) = sin x cos y + cos x sin y)
= 3/4 sin 4x
= RHS
Thus proved.
