 GFG App
Open App Browser
Continue

Half Angle formulas are used to find various values of trigonometric angles such as for 15°, 75°, and others, they are also used to solve various trigonometric problems.

Several trigonometric ratios and identities help in solving problems of trigonometry. The values of trigonometric angles 0°, 30°, 45°, 60°, 90°, and 180° for sin, cos, tan, cosec, sec, and cot are determined using a trigonometry table. Half-Angle formulas are widely used in mathematics, let’s learn about them in detail in this article.

## Half-Angle Formulae

For finding the values of angles apart from the well-known values of 0°, 30°, 45°, 60°, 90°, and 180°. Half angles are derived from double angle formulas and are listed below for sin, cos, and tan:

• sin (x/2) = ± [(1 – cos x)/ 2]1/2
• cos (x/2) = ± [(1 + cos x)/ 2]1/2
• tan (x/ 2) = (1 – cos x)/ sin x

Trigonometric identities of double-angle formulas are useful for the derivation of half-angle formulas.

## Half Angle Identities

Half-angle identities for some popular trigonometric functions are,

• Half Angle Formula of Sin,

sin A/2 = ±√[(1 – cos A) / 2]

• Half Angle Formula of Cos,

cos A/2 = ±√[(1 + cos A) / 2]

• Half Angle Formula of Tan,

tan A/2 = ±√[1 – cos A] / [1 + cos A]

tan A/2 = sin A / (1 + cos A)

tan A/2 = (1 – cos A) / sin A

## Half Angle Formulas Derivation Using Double Angle Formulas

Half-Angle formulas are derived using double-angle formulas. Before learning about half-angle formulas we must learn about Double-angle in Trigonometry, most commonly used double-angle formulas in trigonometry are:

• sin 2x = 2 sin x cos x
• cos 2x = cos2 x – sin2 x
= 1 – 2 sin2x
= 2 cos2x – 1
• tan 2x = 2 tan x / (1 – tan2x)

Now replacing x with x/2 on both sides in the above formulas we get

• sin x = 2 sin(x/2) cos(x/2)
• cos x = cos2 (x/2) – sin2 (x/2)
= 1 – 2 sin2 (x/2)
= 2 cos2(x/2) – 1
• tan A = 2 tan (x/2) / [1 – tan2(x/2)]

### Half-Angle Formula for Cos Derivation

We use cos2x = 2cos2x – 1 for finding the Half-Angle Formula for Cos

Put x = 2y in the above formula

cos (2)(y/2) = 2cos2(y/2) – 1

cos y = 2cos2(y/2) – 1

1 + cos y = 2cos2(y/2)

2cos2(y/2) = 1 + cosy

cos2(y/2) = (1+ cosy)/2

cos(y/2) = ± √{(1+ cosy)/2}

### Half-Angle Formula for Sin Derivation

We use cos 2x = 1 – 2sin2x for finding the Half-Angle Formula for Sin

Put x = 2y in the above formula

cos (2)(y/2) = 1 – 2sin2(y/2)

cos y = 1 – 2sin2(y/2)

2sin2(y/2) = 1 – cosy

sin2(y/2) = (1 – cosy)/2

sin(y/2) = ± √{(1 – cosy)/2}

### Half-Angle Formula for Tan Derivation

We know that tan x  = sin x / cos x such that,

tan(x/2) = sin(x/2) / cos(x/2)

Putting the values of half angle for sin and cos. We get,

tan(x/2) = ± [(√(1 – cosy)/2 ) / (√(1+ cosy)/2 )]

tan(x/2) = ± [√(1 – cosy)/(1+ cosy) ]

Rationalising the denominator

tan(x/2) = ± (√(1 – cosy)(1 – cosy)/(1+ cosy)(1 – cosy))

tan(x/2) = ± (√(1 – cosy)2/(1 – cos2y))

tan(x/2) = ± [√{(1 – cosy)2/( sin2y)}]

tan(x/2) = (1 – cosy)/( siny)

Also, Check

## Solved Examples on Half Angle Formulas

Example 1: Determine the value of sin 15°

Solution:

We know that the formula for half angle of sine is given by:

sin x/2 = ± ((1 – cos x)/ 2) 1/2

The value of sine 15° can be found by substituting x as 30° in the above formula

sin 30°/2 = ± ((1 – cos 30°)/ 2) 1/2

sin 15° = ± ((1 – 0.866)/ 2) 1/2

sin 15° = ± (0.134/ 2) 1/2

sin 15° = ± (0.067) 1/2

sin 15° = ± 0.2588

Example 2: Determine the value of sin 22.5°

Solution:

We know that the formula for half angle of sine is given by:

sin x/2 = ± ((1 – cos x)/ 2) 1/2

The value of sine 15° can be found by substituting x as 45° in the above formula

sin 45°/2 = ± ((1 – cos 45°)/ 2) 1/2

sin 22.5° = ± ((1 – 0.707)/ 2) 1/2

sin 22.5° = ± (0.293/ 2) 1/2

sin 22.5° = ± (0.146) 1/2

sin 22.5° = ± 0.382

Example 3: Determine the value of tan 15°

Solution:

We know that the formula for half angle of sine is given by:

tan x/2 = ± (1 – cos x)/ sin x

The value of tan 15° can be found by substituting x as 30° in the above formula

tan 30°/2 = ± (1 – cos 30°)/ sin 30°

tan 15° = ± (1 – 0.866)/ sin 30

tan 15° = ± (0.134)/ 0.5

tan 15° = ± 0.268

Example 4: Determine the value of tan 22.5°

Solution:

We know that the formula for half angle of sine is given by:

tan x/2 = ± (1 – cos x)/ sin x

The value of tan 22.5° can be found by substituting x as 45° in the above formula

tan 30°/2 = ± (1 – cos 45°)/ sin 45°

tan 22.5° = ± (1 – 0.707)/ sin 45°

tan 22.5° = ± (0.293)/ 0.707

tan 22.5° = ± 0.414

Example 5: Determine the value of cos 15°

Solution:

We know that the formula for half angle of sine is given by:

cos x/2 = ± ((1 + cos x)/ 2) 1/2

The value of sine 15° can be found by substituting x as 30° in the above formula

cos 30°/2 = ± ((1 + cos 30°)/ 2) 1/2

cos 15° = ± ((1 + 0.866)/ 2) 1/2

cos 15° = ± (1.866/ 2) 1/2

cos 15° = ± (0.933) 1/2

cos 15° = ± 0.965

Example 6: Determine the value of cos 22.5°

Solution:

We know that the formula for half angle of sine is given by:

cos x/2 = ± ((1 + cos x)/ 2) 1/2

The value of sine 15° can be found by substituting x as 45° in the above formula

cos 45°/2 = ± ((1 + cos 45°)/ 2) 1/2

cos 22.5° = ± ((1 + 0.707)/ 2) 1/2

cos 22.5° = ± (1.707/ 2) 1/2

cos 22.5° = ± ( 0.853 ) 1/2

cos 22.5° = ± 0.923

## FAQs on Half-Angle Formula

### Question 1: What is the use of Half-Angle Formulas?

Half-Angle formulas are used for finding trigonometric ratios of half of the standard angles such as 15°,22.5° and others. They are also used for solving complex trigonometric equations and are required in solving integrals, and differential equations.

### Question 2: What is Half Angle Formula for Sin?

Half-Angle formula for sin is

sin A/2 = ±√[(1 – cos A) / 2]

Also, for any triangle with sides a, b, and c and semiperimeter be s, then

sin A/2 = √[(s – b) (s – c) / bc]

### Question 3: What is Half Angle Formula for Cosine?

Half-angle formula for cos is

cos A/2 = ±√[(1 + cos A)/2]

Also, for any triangle with sides a, b, and c and semiperimeter be s, then

cos (A/2) = √[ s (s – a)/bc]