Half Angle Formula

Half Angle Formula

Edited By Komal Miglani | Updated on Oct 12, 2024 12:29 PM IST

The half-angle formula is used to find the value of the trigonometric ratios like 22.5°, 15°. half-angle of trigonometric functions with the help of an angle. These formulae can be derived from the reduction formulas and we can use them when we have an angle that is half the size of a special angle. It is used to find the exact value of the trigonometric ratios of 15 (half of 30 degrees), 22.5( half of 45 degrees), and so on.

In this article, we will cover the concept of the half-angle formula. This concept falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Half Angle Formula Definition

The half-angle formula is used to find the value of the trigonometric ratios of the angles like 22.5° (which is half of the angle 45°), 15° (which is half of the angle 30°), etc.

Half Angle Formulas

The half-angle formula can be derived with the help of the reduction formula.
1. sin(α2)=±1cosα2
2. cos(α2)=±1+cosα2
3. tan(α2)=±1cosα1+cosα

Note that the half-angle formulas are preceded by a ± sign. This does not mean that both positive and negative expressions are valid. Rather, only one sign needs to be taken and that depends on the quadrant in which α/2 lies.

Derivation of Half Angle Formulas
The half-angle formula can be derived with the help of the reduction formula.
Reduction formulas are:
sin2θ=1cos(2θ)2
cos2θ=1+cos(2θ)2
tan2θ=1cos(2θ)1+cos(2θ)

Derivation of Half Angle Formula for Sine

The half-angle formula for sine is derived as follows:

sin2θ=1cos(2θ)2sin2(α2)=1cos(2α2)2=1cosα2sin(α2)=±1cosα2

Derivation of Half Angle Formula for Cosine

To derive the half-angle formula for cosine, we have

cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2α2)2=1+cosα2cos(α2)=±1+cosα2

Derivation of Half Angle Formula for Tangent

For the tangent identity, we have

tan2θ=1cos(2θ)1+cos(2θ)tan2(α2)=1cos(2α2)1+cos(2α2)=1cosα1+cosαtan(α2)=±1cosα1+cosα

Derivation of Half Angle Formula Using Semiperimeter

We can also represent the half-angle formula using the side of the triangle.

Derivation of Half-Angle Formula for Sine

sinA2=(sb)(sc)bcsinB2=(sa)(sc)acsinC2=(sa)(sb)ab
We know that,

cosA=12sin2A2sin2A2=1cosA2
Now, for any ABC,cosA=b2+c2a22bc

Using the above two formulas

In a similar way, we can derive other formulas.

sin2A2=12[1b2+c2a22bc]=12[2bcb2c2+a22bc]=12[a2(bc)22bc]=(ab+c)(a+bc)4bc=(2 s2 b)(2 s2c)4bc[a+b+c=2 s]sin2A2=(sb)(sc)bc As 0<A2<π2, so sinA2>0sinA2=(sb)(sc)bc

Derivation of Half-Angle Formula for Cosine

cosA2=(s)(sa)bccosB2=(s)(sb)accosC2=(s)(sc)ab
We know that

cosA=2cos2A21cos2A2=1+cosA2 And for any ABC,cosA=b2+c2a22bc

Using the above two formulas

cos2A2=12[1+b2+c2a22bc]=12[2bc+b2+c2a22bc]=12[(b+c)2a2]2bc]=(b+c+a)(b+ca)4bc=(b+c+a)(a+b+c2a)4bc=(2 s)(2 s2a)4bc[a+b+c=2 s]cos2A2=(s)(sa)bc

As 0<A2<π2, so cosA2>0

cosA2=(s)(sa)bc

In a similar way, we can derive other formulas

Derivation of Half Angle Formula for tan

This half-angle formula can be proved using tan x = sin x/cos x, and using the half-angle formula of sine and cosine.

tanA2=(sb)(sc)s(sa)tanB2=(sa)(sc)s(sb)tanC2=(sa)(sb)s(sc)

This half-angle formula can be proved using $\tan x=\sin x / \cos x and using the half-angle formula of sine and cosine.

Summary

The half-angle formulas in trigonometry serve as crucial tools for simplifying and relating trigonometric functions of angles halved to those of their originals. These formulas have applications in various fields including mathematics, physics, and engineering. Understanding these formulas helps us to solve theoretical and practical problems.

Recommended Video :

Solved Example Based on Half Angle Formula

Example 1: If tan(π9),x,tan(7π18) are in arithmetic progression and tan(π9),y,tan(5π18) are also in arithmetic progression, then |x2y| is equal to [JEE MAINS 2021]
Solution

x=tan(π9)+tan(7π18)2 and y=tan(π9)+tan(5π18)2
x2y=12[tan(π9)+tan(7π18)2tan(π9)2tan(5π18)]=12[tan(7π18)tan(π9)2tan(5π18)]=12[tan(π2π9)tan(π9)2tan(5π18)]=12[cot(π9)tan(π9)2tan(π22π9)]=12[cos(π9)sin(π9)sin(π9)cos(π9)2cot(2π9)]=12[2cos(2π/9)sin(2π9)2cot(2π9)]=0

Hence, the answer is 0

Example 2: The value of 2sin(π8)sin(2π8)sin(3π8)sin(5π8)sin(6π8)sin(7π8) is :
[JEE MAINS 2021]
Solution:

As sin(5π8)=sin(π3π8)=sin(3π8) similarly sin(6π8)=sin(2π8) and sin(7π8)=sin(π8)

Required value

=2sin2(π8)sin2(2π8)sin2(3π8)=2(12)2sin2(π8)sin2(3π8)=sin2(π8)cos2(π8)=44sin2(π8)cos2(π8)=14(sin(π4))2=1412=18
Hence, the answer is 1/8

Example 3: If cosα+cosβ=32 and sinα+sinβ=12 and θ is the arithmetic mean of α and β, then sin2θ+cos2θ is equal to : [JEE MAINS 2015]
Solution:

Trigonometric Ratios of Submultiples of an Angle -

tan2θ=2tanθ1tan2θsinC+sinD=2sin(C+D2)cos(CD2)cosC+cosD=2cos(C+D2)cos(CD2)
This shows the transformation formulae and double angle formulae.
Therefore,

cosα+cosβ=322cosα+β2cosαβ2=32sinα+sinβ=122sinα+β2cosαβ2=12

dividing (2) by (1), we get

tan(α+β)2=13tanθ=13

tan2θ=2×13119=34sin2θ=35,cos2θ=45

[By making a triangle]
Thus,

sin2θ+cos2θ=75
Hence, the answer is 7/5

Example 4: Let α,β be such that π<αβ<3π. If sinα+sinβ=21/65, and
cosα+cosβ=27/65 then the value of cosαβ2 is :

Solution:

sinα+sinβ=21652sin(α+β)2cos(αβ)2=2165

cosα+cosβ=27652cos(α+β)2cos(αβ)2=2165
Squaring and adding,

4cos2(αβ)2[sin2α+β2+cos2α+β2]=212+272652cos2(αβ)2=11704×4225cos2(αβ)2=11704×4225=9130cos(αβ)2=3130[π<αβ<3ππ2αβ23π2cos(αβ2)<0] Hence, the answer is 3130

Example 5: Find the range of functions tanα2tanα
Solution:

tanα2tanα=(1cosαsinα)sinαcosα=secα1 Range of secα(,1]U[1,) Range of f(α) is (,2]U[0,)

Hence, the answer is (,2]U[0,)


Articles

Get answers from students and experts
Back to top