# Multiple Angle Formulae

The important trigonometrical ratios of multiple angle formulae are given below:

(i) sin 2A = 2 sin A cos  A

(ii) cos 2A = cos$$^{2}$$ A - sin$$^{2}$$ A

(iii) cos 2A = 2 cos$$^{2}$$ A - 1

(iv) cos 2A = 1 - 2 sin$$^{2}$$ A

(v) 1 + cos 2A = 2 cos$$^{2}$$ A

(vi) 1 - cos 2A = 2 sin$$^{2}$$ A

(vii) tan$$^{2}$$ A = $$\frac{1 - cos 2A}{1 + cos 2A}$$

(viii) sin 2A = $$\frac{2 tan A}{1 + tan^{2} A}$$

(ix) cos 2A = $$\frac{1 - tan^{2} A}{1 + tan^{2} A}$$

(x) tan 2A = $$\frac{2 tan A}{1 - tan^{2} A}$$

(xi) sin 3A = 3 sin A - 4 sin$$^{3}$$ A

(xii) cos 3A = 4 cos$$^{3}$$ A - 3 cos A

(xiii) tan 3A = $$\frac{3 tan A - tan^{3} A}{1 - 3 tan^{2} A}$$

Now we will learn how to use the above formulae for solving different types of trigonometric problems on multiple angles.

1. Prove that cos 5x = 16 cos$$^{5}$$ x – 20 cos$$^{3}$$ x + 5 cos x

Solution:

L.H.S. = cos 5x

= cos (2x + 3x)

= cos 2x cos 3x - sin 2x sin 3x

= (2 cos$$^{2}$$ x - 1) (4 cos$$^{3}$$ x - 3 cos x) - 2 sin x cos x (3 sin x - 4 sin$$^{3}$$ x)

= 8 cos$$^{5}$$ x - 10 cos$$^{3}$$ x + 3 cos x - 6 cos x sin$$^{2}$$ x + 8 cos x sin$$^{4}$$ x

= 8 cos$$^{5}$$ x - 10 cos$$^{3}$$ x + 3 cos x - 6 cos x (1 - cos$$^{2}$$ x) + 8 cos x (1 - cos$$^{2}$$ x)$$^{2}$$

= 8 cos$$^{5}$$ x - 10 cos$$^{3}$$ x + 3 cos x - 6 cos x + 6 cos$$^{3}$$ x + 8 cos x - 16 cos$$^{3}$$ x + 8 cos$$^{5}$$ x

= 16 cos$$^{5}$$ x - 20 cos$$^{3}$$ x + 5 cos x

2. If 13x = π, proved that cos x cos 2x cos 3x cos 4x cos 5x cos 6x = ½^6

Solution:

L. H. S = cos x cos 2x cos 3x cos 4x cos 5x cos 6x

= $$\frac{1}{2 sin x}$$ (2 sin x cos x) cos 2x cos 3x cos 4x cos 5x  cos 6x

= $$\frac{1}{2 sin x}$$ sin 2x cos 2x cos 3x cos 4x cos 5x cos 6x

= $$\frac{1}{2^2 sin x}$$ (2 sin 2x cos 2x) cos 3x cos 4x cos 5x cos 6x

= $$\frac{1}{2^3 sin x}$$ (2 sin 4x cos 4x) cos 3x cos 5x cos 6x

= $$\frac{1}{2^3 sin x}$$ sin 8x cos 3x cos 5x cos 6x

= $$\frac{1}{2^4 sin x}$$ (2 sin 5x cos 5x) cos 3x cos 6x,

[Since, sin 8x = sin (13x - 5x) = sin (π - 5x), (given 13x = π)

= sin 5x]

= $$\frac{1}{2^4 sin x}$$ sin 10x cos 3x cos 6x

= $$\frac{1}{2^5 sin x}$$ (2 sin 3x cos 3x) cos 6x,

[Since, sin 10x = sin (13x – 3x) = sin (π – 3x), (given 13x = π)

= sin 3x]

= $$\frac{1}{2^6 sin x}$$ 2 sin 3x cos 6x

= $$\frac{1}{2^6 sin x}$$ sin 12x

= $$\frac{1}{2^6 sin x}$$ sin (13x - x)

= $$\frac{1}{2^6 sin x}$$ sin (π - x), [Since, 13x = π]

= $$\frac{1}{2^6 sin x}$$ sin x

= $$\frac{1}{2^6}$$ = R.H.S.                         Proved