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Half Angle formula
Students study the Half Angle Formula(or halfangle identities) in Trigonometry. The halfwidth formula can be derived using the doublewidth formula. As students know, the double angle formula can be derived from the sum of angles and difference of angles formulas in trigonometry. Half Angle Formula in halfwidth expressions is usually expressed as θ/2, x/2, A/2, etc., where halfwidths are partial angles. Halfangle formulas are used to find the exact value of trigonometric ratios for angles such as 22.5° (half the standard 45° angle), 15° (half the standard 30° angle), and so on. Students shall examine the halfwidth formula and its proof, as well as some solved examples.
Quick Links
ToggleWhat are Half Angle Formulas?
In this section, students will look at Half Angle Formula for sin, cos, and tan. From the table of trigonometric functions, know the values of trigonometric functions (sin, cos, tan, cot, sec, cosec) for angles such as 0°, 30°, 45°, 60°, 90°. However, to know the exact values of sin 22.5°, tan 15°, etc., halfangle formulas are very useful. It is also useful for proving multiple trigonometric identities. There are halfwidth formulas derived from the doublewidth formula, expressed in halfwidth θ/2, x/2, A/2, and so on. Below is a list of important Half Angle Formula.
These are Greek letters such as alpha (α), beta (β), theta (θ), and gamma (γ) that represent angles. These represent the sign of the trigonometric function in each quadrant. Several different units of angular measurement are commonly used, such as degrees, radians, and grads. As an example, one full circle (rotation) = 360° = 2π radians = 400 grads. However, all angle values are given in radians, not specifically marked with (°) for degrees or (g) for grads.
Half Angle Identities
Here are some common halfangle identities and Half Angle Formula that students use to solve many trigonometric problems:
 Formula of sin: sin A/2 = ±√[(1 – cos A) / 2]
 Formula of cos: cos A/2 = ±√[(1 + cos A) / 2]
 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
Derivation of Half Angle Formula shall occur by using the doubleangle formula.
To derive the above formula, one must first derive the following Half Angle Formula: Double angle formulas represent double angles such as 2θ, 2A, and 2x. The doubleangle formulas are known to be sin, cos, and tan.
sin 2x = 2 sin x cos x
cos 2x = cos2 x – sin2 x (or)
= 1 – 2 sin2x (or)
= 2 cos2x – 1
tan 2x = 2 tan x / (1 – tan2x)
Replacing x with A/2 on both sides of each equation in the doubleangle formula gives the halfangle identity (as 2x = 2(A/2) = A).
sin A = 2 sin(A/2) cos(A/2)
cos A = cos2 (A/2) – sin2 (A/2) (or)
= 1 – 2 sin2 (A/2) (or)
= 2 cos2(A/2) – 1
tan A = 2 tan (A/2) / (1 – tan2(A/2))
Students can also derive a Half Angle Formula using another Half Angle Formula. For example, from the cos A 3 equation alone, students can derive the important Half Angle Formula identities for sin, cos, and tan mentioned in the first section. Here is the proof of the Half Angle Formula.
Half Angle Formula of Sin Proof
Next, students prove the Half Angle Formula for the sine function. Using either of the cos A formulas above, students get:
cos A = 1 – 2 sin2 (A/2)
From this,
2 sin2 (A/2) = 1 – cos A
sin2 (A/2) = (1 – cos A) / 2
sin (A/2) = ±√[(1 – cos A) / 2]
Half Angle Formula of Cos Derivation
Here students have to prove the Half Angle Formula for the cosine function. Using either of the cos A expressions above,
cos A = 2 cos2(A/2) – 1
From this,
2 cos2(A/2) = 1 + cos A
cos2 (A/2) = (1 + cos A) / 2
cos (A/2) = ±√[(1 + cos A) / 2]
Half Angle Formula of Tan Derivation
tan(A/2) = [sin(A/2)] / [cos(A/2)]
From the Half Angle Formula 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 expressions for tan (A/2). Now students must rationalise the denominator and derive the remaining two equations.
tan (A/2) = ±√[(1 – cos A) / (1 + cos A)] × √[(1 – cos A) / (1 – cos A)]
= √[(1 – cos A)2 / (1 – cos2A)]
= √[(1 – cos A)2/ sin2A]
= (1 – cos A) / sin A
This is the second expression for tan (A/2). To derive another formula, students must multiply and divide the above formula by (1 + cos A). Then they would get
tan (A/2) = [(1 – cos A) / sin A] × [(1 + cos A) / (1 + cos A)]
= (1 – cos2A) / [sin A (1 + cos A)]
= sin2A / [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, students will look at the Half Angle Formula using the perimeter of halfcircles. So these are the Half Angle Formula for the sides of a triangle. Consider triangle ABC with AB = c, BC = a, and CA = b.
Students must read how to derive one of these formulas here. Students know that the semiperimeter of the triangle is s = (a + b + c)/2. From this, students 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
using the law of cosines now,
2 cos2(A/2) = 1 + [ (b2 + c2 – a2) / (2bc) ]
2 cos2(A/2) = [2bc + b² + c² – a²] / [2bc]
2 cos2(A/2) = [ (b + c)² – a²] / [2bc] [Using (a+b)² formula]
2 cos2(A/2) = [ (b + c + a) (b + c – a) ] / [2bc] [Using a² – b² formula]
2 cos2(A/2) = [ 2s (2s – 2a) ] / [2bc] [As 2s = a + b + c]
2 cos2(A/2) = [ 2s (s – a) ] / [bc]
cos2(A/2) = [ s(s – a) ] / [bc]
cos (A/2) = √[ s (s – a) ] / [bc]
Students have derived the Half Angle Formula for the cosine of the angle A/2. Similarly, the halfcircle can be used to derive other halfangle identities for cosine. Another Half Angle Formula for the sine can be derived using the halfcircumference.
sin2(A/2) = (1 − cos A)/2
= (1/2)[1−(b2+c2−a2)/2bc] (Using the law of cosines)
= (1/2)(a2−(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, students can derive other Half Angle Formula for the sine function. The Half Angle Formula for the tangent function can be derived using the formula tan(A/2) = sin(A/2)/cos(A/2).
Note that in the sine and cosine halfangle problems, each square root (root) is preceded by a plus/minus sign. Whether the answer is negative or positive depends on which quadrant the new half angle is in. The above does not apply to tangents, since the Half Angle Formula does not have a preceding ± sign. The reference materials available on the Extramarks website could prove to be very helpful for students to understand the concept of Half Angle Formula. All important Half Angle Formula and their conversions are available. Solved examples help students apply Half Angle Formula to problems.
Examples Using Half Angle Formula
 Example 1: Find the exact value of cos π/8 using the appropriate Half Angle Formula
Solution:
Using the Half Angle Formula for cos,
cos A/2 = ±√[(1 + cos A )/ 2]
Students know that π/8 = 22.5°.
Replace A = 45° on both sides.
cos 45°/2 = ±√[(1 + cos 45°) / 2]
From the ternary plot, students can see 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
However, 22.5° is in quadrant I, so cos 22.5° is positive. therefore,
cos 22.5° = √(2 + √2) / 2
Solution: cos π/8 = √(2 + √2) /
 Example 2: Prove that cos A / (1 + sin A) = tan [ (π/4) – (A/2) ].
Solution:
LHS = cos x / (1 + sin x)
Using Half Angle Formula
= [cos2(A/2) – sin2(A/2)] / [1 + 2 sin(A/2) cos(A/2)]
students know that 1 = cos2(A/2) + sin2(A/2). yes
= [ (cos(A/2) + sin(A/2)) (cos(A/2) – sin(A/2)) ] / [cos2(A/2) + sin2(A/2) + none (A/2) cos (A/2)]
= [ (cos(A/2) + sin(A/2)) (cos(A/2) – sin(A/2)) ] / [cos(A/2) + sin(A/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))
Students know that 1 = tan (π/4).
= (If (π/4) – If (A/2)) / (1 + If (π/4) (A/2))
If (A – If B) / (1 + If A B) = (A – B). yes
= tan [ (π/4) – (A/2)
= RHS
Hence, proved.
 Example 3: For triangle ABC, where AB = c = 12, BC = a = 13, CA = b = 5, find the value of sin A/2.
Solution:
Then a = 13; b = 5; c = 12.
Then the half circumference is s = (a + b + c) / 2 = (13 + 5 + 12) / 2 = 15.
Using the halfangle identity of sin with respect to the halfcircle,
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.
Half Angle Formulas Problems
 What is the Half Angle Formula?
Trigonometry uses Half Angle Formula to determine the exact values of halfangles, such as 15°, 22.5°, and so on.
2. What is the Half Angle Formula for the sine function?
The Half Angle Formula for the sine function is sin a/2 = ±√[(1 – cos a) / 2].
3. What is the Half Angle Formula for the cosine function?
The Half Angle Formula for the cosine function is cos a/2 = ±√[(1 + cos a) / 2].
4. What is the Half Angle Formula for the tangent function?
The Half Angle Formula for the tangent function is tan a/2 = ±√[1 – cos a] / [1 + cos a].
FAQs (Frequently Asked Questions)
1. What is the Half Angle Formula in Trigonometry?
The Half Angle Formula gives halfangle values like A/2, x/2, etc. in trigonometric ratios. The Half Angle Formula for 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]
2. What is the halfangle formula for sin?
The halfangle formula for sin in trigonometry is sin A/2 = ±√[(1 – cos A) / 2]. There is another halfangle formula for sin for the halfcircle. If a, b, and c are the sides of a triangle and A, B, and C are the corresponding diagonals, then sin A/2 = √[(s – b)(s – c)/bc].
3. What is the halfwidth formula for cosine?
The halfwidth formula for cos is cos A/2 = ±√[(1 + cos A)/2]. There is another halfangle formula for cos with respect to the halfcircle. If a, b, and c are the sides of a triangle and A, B, and C are the corresponding diagonals, then cos(A/2) = √[ s(s – a)/bc].
4. What is meant by the tangent halfangle formula?
The tangent halfangle formula is tan (A/2) = ±√[1 – cos A] / [1 + cos A] = (1 – cos A) / sin A = sin A / (1 + cos A ). There is another halfangle formula for tan for the halfcircle. If a, b, and c are the sides of a triangle and A, B, and C are the corresponding diagonals, then sin A/2 = √[(s – b) (s – c) ] / [s( sa)]. Reasons for using halfwidth expressions Use the halfangle formula to find the trigonometric ratio of half the standard angle. For example, one can use the halfangle identity to find trigonometric ratios of angles such as 15°, 22.5°, and so on. They can be used to prove identities for various trigonometric functions. It is also used to solve integrals.