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

● Multiple Angles

• sin 2A in Terms of A
• cos 2A in Terms of A
• tan 2A in Terms of A
• sin 2A in Terms of tan A
• cos 2A in Terms of tan A
• Trigonometric Functions of A in Terms of cos 2A
  
• sin 3A in Terms of A
• cos 3A in Terms of A
• tan 3A in Terms of A
• Multiple Angle Formulae

11 and 12 Grade Math

From Multiple Angle Formulae to HOME PAGE