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# Multiple Angle Formulas

• Last Updated : 17 Jun, 2022

Trigonometry is one of the important topics in mathematics that is used in various fields. The trigonometric formulae are applied and used in various formulae, derivations, etc. This article is about the multiple angle formulae in trigonometry where we find sine, cosine, and tangent for multiple angles. This formula can easily evaluate the multiple angles for any given problem. The trigonometric functions with multiple angles are called the multiple angle formulas. Double, half and triple angles are present under multiple angles.

### Multiple Angle Formulae

When the trigonometric functions with multiple angles are called multiple angle formula. Double, triple and half angle formulae are under multiple angle formulae.

• Half angle formulae: Consist of the half angle of trigonometric functions.
• Double angle formulae: Consist of the double angle of trigonometric functions.
• Triple angle formulae: Consist of the triple angle of trigonometric functions.

### List of Multiple Angle Formulae

• Half Angle Formulae:
• sin x = 2 sin(x/2)cos(x/2) = (2 tan (x/2))/(1 + tan2(x/2))
• cos x = cos2(x/2) – sin2(x/2) = 2cos2(x/2) – 1 = 1 – 2sin2(x/2) = (1 – tan2(x/2))/(1 + tan2(x/2))
• tan x = (2 tan(x/2))/(1 – tan2(x/2))
• sin(x/2) = √((1 – cos x)/2)
• cos(x/2) = √((1 + cos x)/2)
• tan(x/2) = √((1 – cos x)/(1 + cos x))
• Double Angle Formulae:
• sin 2x = 2sin x cos x = (2 tan x)/(1 + tan 2x)
• cos 2x = cos 2x – sin 2x = 2cos 2x – 1 = 1 – 2sin 2x = (1 – tan 2x)/(1 + tan 2x)
• tan 2x = (2 tan x)/(1 – tan 2x)
• Triple Angle Formulae:
• sin 3x = 3sin x – 4sin3x
• cos 3x = 4cos3x – 3cos x
• tan 3x = (3tan x – tan3x)/(1 – 3tan2x)

### Generalized Multiple Angle Formulae

• • • ### Sample Problems

Problem 1: Evaluate the following in the terms of tan: (i) sin 8x and (ii) cos 6x.

Solution:

(i) sin 8x = sin (2(4x)) [Since, sin 2x = 2tan x /(1 + tan2x)]

= (2tan 4x)/(1+tan2(4x))

(ii) cos 6x = cos(2(3x)) [Since, cos 2x = (1 – tan2x)/(1 + tan2x)]

= (1 – tan23x)/(1 + tan23x)

Problem 2: Evaluate the following in terms of its half-angle: (i) sin 6x, (ii) cos 8x, and (iii) tan 12x.

Solution:

(i) sin 6x = 2 sin 3x cos 3x [Using half angle formulae]

Or

= 2tan 3x /(1 + tan23x)

(ii) cos 8x = cos24x – sin24x [Using half angle formulae]

Or

cos 8x = 1 – 2sin24x

Or

cos 8x = 2cos24x -1

Or

cos 8x = (1 – tan24x)/(1 + tan24x)

(iii) tan 12x = (2tan 6x) / (1 + tan26x)

Problem 3: Find cos 4x in terms of cos x.

Solution:

cos 4x = cos (2(2x))

= 2 cos22x – 1

= 2×(2cos2x – 1)2 – 1

= 2×(4cos4x + 1 – 4cos2x) – 1

= 8cos4x + 2 – 8cos2x – 1

= 8cos4x – 8cos2x + 1

Problem 4: If sin 2x = 4/5, find the value of tan x.

Solution:

sin 2x = 4/5

cos 2x = √1 – sin2 2x = √1 – (4/5)2 = √1 – 16/25 = √9/25 = 3/5

As, tan x = √(1 – cos 2x)/(1 + cos 2x)

tan x = √(1 – (3/5))/(1 + (3/5))

tan x = √(2/5)/(8/5)

tan x = √1/4 = 1/2

Problem 5: If cot x = p, then find the value of sec 2x – tan 2x in terms of p.

Solution:

sec 2x – tan 2x = (1/cos 2x) – (sin 2x/cos 2x)

= (1 – sin 2x)/cos 2x [Using double angle formulae and sin2x + cos2x = 1]

= (sin2x + cos2x – 2sin x cos x)/ (cos2x – sin2x)

= (cos x – sin x)2 /(cos x – sin x)(cos x + sin x)

= (cos x – sin x) /(cos x + sin x)

Dividing both numerator and denominator by sin x

sec 2x – tan 2x = (cot x – 1)/(cot x + 1)

sec 2x – tan 2x = (p – 1)/(p + 1)

Problem 6: Find the value of 12 sin 10° – 16 sin310°.

Solution:

12 sin 10° -16 sin310° = 4 [3sin10° – 4sin310°]

= 4[sin 3× 10°] [Since, sin 3A = 3sin A – 4sin3A]

= 4 sin 30°

= 4 × (1/2)

= 2

Problem 7: Prove that – (cot A + cosec A)2 + 1 = cosec2(A/2).

Solution:

LHS = (cot A + cosec A)2 + 1

= {(cos A /sin A) + (1/sin A)}2 + 1

= [(1 + cos A)/ sin A]2 + 1

= [2 cos2(A/2)/ 2 sin(A/2) cos(A/2)]2 + 1

= [cos (A/2) / sin(A/2)]2 +1

= cot2(A/2) + 1

= cosec2(A/2)

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