Half Angle Formulas
We study half angle formulas (or halfangle identities) in Trigonometry. Half angle formulas can be derived using the double angle formulas. As we know, the double angle formulas can be derived using the angle sum and difference formulas of trigonometry. Halfangles in half angle formulas are usually denoted by θ/2, x/2, A/2, etc and the halfangle is a submultiple angle. The half angle formulas are used to find the exact values of the trigonometric ratios of the angles like 22.5° (which is half of the standard angle 45°), 15° (which is half of the standard angle 30°), etc.
Let us explore the half angle formulas along with their proofs and with a few solved examples here.
What are Half Angle Formulas?
In this section, we will see the half angle formulas of sin, cos, and tan. We know the values of the trigonometric functions (sin, cos , tan, cot, sec, cosec) for the angles like 0°, 30°, 45°, 60°, and 90° from the trigonometric table. But to know the exact values of sin 22.5°, tan 15°, etc, the half angle formulas are extremely useful. Also, they are helpful in proving several trigonometric identities. We have half angle formulas that are derived from the double angle formulas and they are expressed in terms of half angles like θ/2, x/2, A/2, etc. Here is the list of important half angle formulas:
Half Angle Identities
Here are the popular half angle identities that we use in solving many trigonometry problems are as follows:
 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] (or) sin A / (1 + cos A) (or) (1  cos A) / sin A
Half Angle Formulas Derivation Using Double Angle Formulas
To derive the above formulas, first, let us derive the following half angle formulas. The double angle formulas are in terms of the double angles like 2θ, 2A, 2x, etc. We know that the double angle formulas of sin, cos, and tan are
 sin 2x = 2 sin x cos x
 cos 2x = cos^{2} x  sin^{2} x (or)
= 1  2 sin^{2}x (or)
= 2 cos^{2}x  1  tan 2x = 2 tan x / (1  tan^{2}x)
If we replace x with A/2 on both sides of every equation of double angle formulas, we get half angle identities (as 2x = 2(A/2) = A).
 sin A = 2 sin(A/2) cos(A/2)
 cos A = cos^{2} (A/2)  sin^{2} (A/2) (or)
= 1  2 sin^{2} (A/2) (or)
= 2 cos^{2}(A/2)  1  tan A = 2 tan (A/2) / (1  tan^{2}(A/2))
We can also derive one half angle formula using another half angle formula. For example, just from the formula of cos A, we can derive 3 important half angle identities for sin, cos, and tan which are mentioned in the first section. Here is the half angle formulas proof.
Half Angle Formula of Sin Proof
Now, we will prove the half angle formula for the sine function. Using one of the above formulas of cos A, we have
cos A = 1  2 sin^{2} (A/2)
From this,
2 sin^{2} (A/2) = 1  cos A
sin^{2} (A/2) = (1  cos A) / 2
sin (A/2) = ±√[(1  cos A) / 2]
Half Angle Formula of Cos Derivation
Now, we will prove the half angle formula for the cosine function. Using one of the above formulas of cos A,
cos A = 2 cos^{2}(A/2)  1
From this,
2 cos^{2}(A/2) = 1 + cos A
cos^{2} (A/2) = (1 + cos A) / 2
cos (A/2) = ±√[(1 + cos A) / 2]
Half Angle Formula of Tan Derivation
We know that tan (A/2) = [sin (A/2)] / [cos (A/2)]
From the half angle formulas of sin and cos,
tan (A/2) = [±√(1  cos A)/2] / [±√(1 + cos A)/2]
= ±√[(1  cos A) / (1 + cos A)]
This is one of the formulas of tan (A/2). Let us derive the other two formulas by rationalizing the denominator here.
tan (A/2) = ±√[(1  cos A) / (1 + cos A)] × √[(1  cos A) / (1  cos A)]
= √[(1  cos A)^{2} / (1  cos^{2}A)]
= √[(1  cos A)^{2}/ sin^{2}A]
= (1  cos A) / sin A
This is the second formula of tan (A/2). To derive another formula, let us multiply and divide the above formula by (1 + cos A). Then we get
tan (A/2) = [(1  cos A) / sin A] × [(1 + cos A) / (1 + cos A)]
= (1  cos^{2}A) / [sin A (1 + cos A)]
= sin^{2}A / [sin A (1 + cos A)]
= sin A / (1 + cos A)
Thus, tan (A/2) = ±√[(1  cos A) / (1 + cos A)] = (1  cos A) / sin A = sin A / (1 + cos A).
Half Angle Formula Using Semiperimeter
In this section, we will see the half angle formulas using the semi perimeter. i.e., these are the half angle formulas in terms of sides of a triangle. Let us consider a triangle ABC where AB = c, BC = a, and CA = b.
Let us derive one of these formulas here. We know that the semiperimeter of the triangle is s = (a + b + c)/2. From this, we have 2s = a + b + c. From one of the above formulas,
cos A = 2 cos²(A/2)  1 (or)
2 cos²(A/2) = 1 + cos A
Now using the law of cosines,
2 cos^{2}(A/2) = 1 + [ (b^{2} + c^{2}  a^{2}) / (2bc) ]
2 cos^{2}(A/2) = [2bc + b² + c²  a²] / [2bc]
2 cos^{2}(A/2) = [ (b + c)²  a²] / [2bc] [Using (a+b)² formula]
2 cos^{2}(A/2) = [ (b + c + a) (b + c  a) ] / [2bc] [Using a²  b² formula]
2 cos^{2}(A/2) = [ 2s (2s  2a) ] / [2bc] [As 2s = a + b + c]
2 cos^{2}(A/2) = [ 2s (s  a) ] / [bc]
cos^{2}(A/2) = [ s(s  a) ] / [bc]
cos (A/2) = √[ s (s  a) ] / [bc]
We have derived one halfangle formula for cosine of angle A/2. Similarly, we can derive other half angle identities of cosine using the semi perimeter. Another half angle formula of sine can be derived using the semi perimeter.
sin^{2}(A/2) = (1 − cos A)/2
= (1/2)[1−(b^{2}+c^{2}−a^{2})/2bc] (Using the law of cosines)
= (1/2)(a^{2}−(b−c)^{2})/2bc
= (1/2)(a + b − c)(a + c − b)/2bc
= (1/2){(a + b + c) − 2c}{(a + b + c) − 2b}/2bc
= (1/2)(2s − 2c)(2s − 2b)/2bc
= (s − b)(s − c)/bc
⇒ sin (A/2) = √[(s − b)(s − c)/bc]
Similarly, we can derive other half angle formulas of the sine function. Half angle formulas for tangent function can be derived using the formula tan (A/2) = sin (A/2)/cos (A/2).
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Examples Using Half Angle Formula

Example 1: Use an appropriate half angle formula to find the exact value of cos π/8.
Solution:
Using the half angle formula of cos,
cos A/2 = ±√[(1 + cos A )/ 2]
We know that π/8 = 22.5°.
Substitute A = 45° on both sides,
cos 45°/2 = ±√[(1 + cos 45°) / 2]
From Trig chart, we know that cos 45° = √2/2.
cos 22.5° = ±√[1 + (√2/2) / 2]
cos 22.5° = ±√[(2 + √2) / (2 × 2)]
cos 22.5° = ± √(2 + √2) / 2
But 22.5° lies in quadrant I and hence cos 22.5° is positive. Thus,
cos 22.5° = √(2 + √2) / 2
Answer: cos π/8 = √(2 + √2) / 2.

Example 2: Prove that cos A / (1 + sin A) = tan [ (π/4)  (A/2) ].
Solution:
LHS = cos x / (1 + sin x)
Using the half angle formulas,
= [cos^{2} (A/2)  sin^{2} (A/2)] / [1 + 2 sin (A/2) cos (A/2)]
We know that 1 = cos^{2}(A/2) + sin^{2}(A/2). So
= [ (cos (A/2) + sin (A/2)) (cos (A/2)  sin (A/2)) ] / [cos^{2}(A/2) + sin^{2}(A/2) + 2 sin (A/2) cos (A/2)]
= [ (cos (A/2) + sin (A/2)) (cos (A/2)  sin (A/2)) ] / [cos (A/2) + sin (A/2)]^{2}
= [cos (A/2)  sin (A/2)] / [cos (A/2) + sin (A/2)]
= [ cos (A/2) ( 1  sin (A/2)/cos(A/2) ) ] / [ cos (A/2) ( 1 + sin (A/2)/cos(A/2) ) ]
= (1  tan (A/2)) / (1 + tan (A/2))
We know that 1 = tan (π/4). So
= (tan (π/4)  tan (A/2)) / (1 + tan (π/4) tan (A/2))
We have (tan A  tan B) / (1 + tan A tan B) = tan (A  B). So
= tan [ (π/4)  (A/2) ]
= RHS
Hence proved.
Answer: We proved the given identity.

Example 3: In a triangle ABC, if AB = c = 12, BC = a = 13, and CA = b = 5, then find the value of sin A/2.
Solution:
It is given that a = 13; b = 5; c = 12.
Then the semiperimeter is, s = (a + b + c) / 2 = (13 + 5 + 12) / 2 = 15.
Using the half angle identity of sin in terms of semi perimeter,
sin A/2 = √[(s  b) (s  c) / bc]
= √[(15  5) (15  12) /(5)(12)]
= √[(10) (3) / 60]
= √2/2
Answer: sin A/2 = √2/2.
FAQs on Half Angle Formula
What Are Half Angle Formulas in Trigonometry?
The half angle formulas give the value of half angles like A/2, x/2, etc of trigonometric ratios. The half angle formulas of sin, cos, and tan are
 sin A/2 = ±√[(1  cos A) / 2]
 cos A/2 = ±√[(1 + cos A) / 2]
 tan A/2 = ±√[1  cos A] / [1 + cos A]
What Is Half Angle Formula for Sin?
The half angle formula of sin in trigonometry is sin A/2 = ±√[(1  cos A) / 2]. We have another halfangle formula of sin in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then sin A/2 = √[(s  b) (s  c)/bc].
What Is Half Angle Formula for Cosine?
The half angle formula of cos is cos A/2 = ±√[(1 + cos A)/2]. We have another half angle formula of cos in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then cos (A/2) = √[ s (s  a)/bc].
What Is Half Angle Formula for Tangent?
The half angle formula of tangent is tan (A/2) = ±√[1  cos A] / [1 + cos A] = (1  cos A) / sin A = sin A / (1 + cos A). We have another half angle formula of tan in terms of semiperimeter. If a, b, and c are the sides of a triangle and A, B, and C are their corresponding opposite angles, then sin A/2 = √[(s  b) (s  c) ] / [s(s  a)].
Why To Use Half Angle Formulas?
We use half angle formulas in finding the trigonometric ratios of the half of the standard angles, for example, we can find the trigonometric ratios of angles like 15°, 22.5°, etc using the half angle identities. They can be used in proving various trigonometric identities. They are also used in solving integrals.
How To Derive Half Angle Formula of Cos?
Using the double angle formula of cos,
cos 2x = 2cos^{2}x  1
By replacing x with (A/2), we get
cos A = 2 cos^{2}(A/2)  1
We will solve this for cos (A/2).
2 cos^{2}(A/2) = 1 + cos A
cos^{2} (A/2) = (1 + cos A) / 2
cos A/2 = ±√(1 + cos A) / 2
What Is tan 15° Using Half Angle Identities?
Using the half angle identity of tan,
tan (A/2) = (1  cos A) / sin A
Substitute A = 30°,
tan (30°/2) = (1  cos 30°) / sin 30°
= [1  (√3/2)] / (1/2) (From Trig table)
= [ (2  √3) / 2] / (1/2)
= 2  √3
Therefore, tan 15° = 2  √3.
What is the Difference Between Double Angle and Half Angle Formulas?
The double angle formulas in trigonometry are:
 sin 2x = 2 sin x cos x
 cos 2x = cos^{2} x  sin^{2} x (or)
= 1  2 sin^{2}x (or)
= 2 cos^{2}x  1  tan 2x = 2 tan x / (1  tan^{2}x)
The half angle formulas (which are derived from the above formulas) in trigonometry are:
 sin A/2 = ±√[(1  cos A) / 2]
 cos A/2 = ±√[(1 + cos A) / 2]
 tan A/2 = ±√[1  cos A] / [1 + cos A]
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