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






11 and 12 Grade Math

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