Semiperimeter And Half Angle Formulae

Semiperimeter And Half Angle Formulae

Edited By Komal Miglani | Updated on Oct 12, 2024 01:10 PM IST

In trigonometry, the half-angle formula is used to find the value of angles apart from 30,45,60,90. It makes our calculation with the help of the half-angle formula we can find the value of any angle. In real life, we use the half-angle formula to calculate the angle of the roof, ceramic tile installation, etc.

Semiperimeter And Half Angle Formulae
Semiperimeter And Half Angle Formulae

In this article, we will cover the concept of the half-angle formula. This category 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. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of eight questions have been asked on this concept.

Half angle formula

The half-angle formula helps us to find the value of angles like 15,22.5, or any other angle of a triangle with the help of the sides. 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 Trignometric Table. But to know the exact values of sin22.5,tan15, etc, the half angle formulas are extremely useful.

Half-Angle Formula for Sine

We have the following half-angle formula for sine in triangle ABC,

sinA2=(sb)(sc)bcsinB2=(sa)(sc)acsinC2=(sa)(sb)ab

Derivation of Half Angle Formula of sine

cosA=12sin2A2sin2A2=1cosA2

Now, for any ABC,

cosA=b2+c2a22bc

Using the above two 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>0

sinA2=(sb)(sc)bc

In a similar way, we can derive other formulas.

Half-Angle Formula for Cosine

We have the following half-angle formula for cosine in triangle ABC

cosA2=(s)(sa)bccosB2=(s)(sb)accosC2=(s)(sc)ab

Derivation of Half Angle Formula of Cosine
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)2a22bc]=(b+c+a)(b+ca)4bc=(b+c+a)(a+b+c2a)4bc
Similarly, we can derive other formulas

Half Angle Formula for tan

We have the following half-angle formulas for tan in triangle ABC,

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

This half-angle formula can be proved using tanx=sinx/cosx, and using the half-angle formula of sine and cosine.

Half Angle Formula

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.

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.


1. sin(α2)=±1cosα2
2. cos(α2)=±1+cosα2
3. tan(α2)=±1cosα1+cosα

Derivation of Half Angle Formula of 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 of Cosine
To derive the half-angle formula for cosine, we have

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

Derivation of Half Angle Formula of Tangent


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

Summary: The half-angle formula helps us to find the value of angles like 15,22.5, or any other angle of a triangle with the help of the sides. Knowledge of the Half Angle formula helps us to solve complex trignometric problems.

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Solved Examples Based on Half-Angle Formula
Example 1: In triangle ABC if tanA2,tanB2,tanC2 are in H.P. then the sides of triangle ABC will be in.

Solution

Half-Angle Formula (in terms of perimeter and sides of the triangle)

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

Hence, the answer is sides of triangle ABC will be in AP

Example 2: If in a triangle ABC,(sa)(sb)=s(sc) then angle C is equal to.

tanA2,tanB2,tanC2 are in H.P. 2tanB2=1tanA2+1tanC22s(sb)(sa)(sc)=s(sa)(sb)(sc)+s(sc)(sa)(sb)2s(sb)(sa)(sc)=s(sa)2+s(sb)2(sa)(sb)(sc)2s(sb)2=s[(sa)2+(sb)2]2(sb)=(sa)+(sc)a+c=2ba,b,c is in A.P.
Solution: Half-Angle Formula (in terms of perimeter and sides of the triangle.

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

tanC2=(sa)(sb)s(sc)(sa)(sb)=s(sc)tanC2=1=tanπ4C=π2

Hence, the answer is angle C is π2

Example 3: The value of tanB2tanC2 if b+c=2a, is.
Solution

tanB2tanC2=(sa)(sc)s(sb)(sa)(sb)s(sc)=sas=b+caa+b+c=13

Hence, the answer is 13

Example 4: In triangle ABC if tanA2=38 and tanC2=89 then sides a,b,c are in
Solution

tanA2tanC2=(sb)(sc)s(sa)(sa)(sb)s(sc)=sbb=a+cba+b+c=39

(since tanA2=38 and tanC2=89 )

3(a+cb)=a+b+ca+c=2b

a,b,c are in A.P.
Hence, the answer is A.P

Example 5: In a ABCcotA2cotB2cotC2 is equal to

Solution: Trigonometric Ratios of Functions -
cosecθ=HypOpp
secθ=Hyp Base
cotθ= Base Opp

- wherein

Trigonometric Ratios of Functions 2
cotA2cotB2cotC2=s(sa)Δs(sb)Δs(sc)Δ=s3(a+b+c)Δ3=s2Δ2Δ3=s2Δ=(Δ2r2)×1Δ=Δr2cotA2cotB2cotC2=Δr2

Hence, the answer is Δr2
Frequently Asked Questions (FAQs)
Q1) what is the half-angle formula for cosine?
Answer: We have the following half-angle formula for cosine in triangle ABC

cosA2=(s)(sa)bccosB2=(s)(sb)accosC2=(s)(sc)ab



Frequently Asked Questions (FAQs)

1. What is the Half-angle formula for tangent?

We have the following half-angle formulas for tan in triangle ABC,

$
\begin{aligned}
& \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\
& \tan \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \\
& \tan \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}
\end{aligned}
$

2. In triangle $A B C$ if $\cos \frac{B}{2}=\sqrt{\frac{a+c}{2 c}}$, then angle C is

$
\begin{aligned}
& \cos \frac{B}{2}=\sqrt{\frac{a+c}{2 c}} \\
& \sqrt{\frac{s(s-b)}{a c}}=\sqrt{\frac{a+c}{2 c}} \quad \text { where } s=\frac{a+b+c}{2} \\
& \frac{s(s-b)}{a c}=\frac{a+c}{2 c} \\
& 2 s(s-b)=a(a+c) \\
& (a+b+c)(a+c-b)=2 a(a+c) \\
& (a+c)^2-b^2=2 a^2+2 a c \\
& a^2+b^2=c^2 \\
& \angle C=\frac{\pi}{2}
\end{aligned}
$

3. What is the half-angle formula for sine?

We have the following half-angle formula for sine in triangle ABC,

$
\begin{aligned}
\sin \frac{A}{2} & =\sqrt{\frac{(s-b)(s-c)}{b c}} \\
\sin \frac{B}{2} & =\sqrt{\frac{(s-a)(s-c)}{a c}} \\
\sin \frac{C}{2} & =\sqrt{\frac{(s-a)(s-b)}{a b}}
\end{aligned}
$

4. What is the formula for calculating the perimeter of a triangle?

Semi-perimeter of the $\triangle A B C$, is

$
\text { is } s=\frac{a+b+c}{2}
$

and it is denoted by $s$. So, the perimeter of $\triangle A B C$ is $2 s=a+b+c$

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