We study half angle formulas (or half-angle 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. Half-angles in half angle formulas are usually denoted by θ/2, x/2, A/2, etc and the half-angle is a sub-multiple 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.
| 1. | What are Half Angle Formulas? |
| 2. | Half Angle Identities |
| 3. | Half Angle Formulas Derivation Using Double Angle Formulas |
| 4. | Half Angle Formula of Sin Proof |
| 5. | Half Angle Formula of Cos Derivation |
| 6. | Half Angle Formula of Tan Derivation |
| 7. | Half Angle Formula Using Semiperimeter |
| 8. | FAQs on Half Angle Formula |
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:
Here are the popular half angle identities that we use in solving many trigonometry problems are as follows:
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
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).
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.
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)
2 sin 2 (A/2) = 1 - cos A
sin 2 (A/2) = (1 - cos A) / 2
sin (A/2) = ±√[(1 - cos A) / 2]
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
2 cos 2 (A/2) = 1 + cos A
cos 2 (A/2) = (1 + cos A) / 2
cos (A/2) = ±√[(1 + cos A) / 2]
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).
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 semi-perimeter 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
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 half-angle 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 + b − c)(a + c − b)/2bc
= (1/2)(2s − 2c)(2s − 2b)/2bc
⇒ 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).
☛ Related Topics:
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.
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