Semiperimeter And Half Angle Formulae

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.

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This Story also Contains

1. Half angle formula
2. Half Angle Formulas Derivation Using Double Angle Formulas
3. Half-Angle Formula for Sine
4. Half-Angle Formula for Cosine
5. Half Angle Formula for tan
6. Half Angle Formula
7. Half Angle Formula Using Semiperimeter
8. Solved Examples Based on Half-Angle Formula

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 $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}$, and $90^{\circ}$ from the Trignometric Table. But to know the exact values of $\sin 22.5^{\circ}, \tan 15^{\circ}$, etc, the half angle formulas are extremely useful.

Half Angle Identities

Here are the popular half angle identities that we use in solving many trigonometry problems are as follows:

• Sine Half-Angle Formula: $
  \sin \frac{A}{2}= \pm \sqrt{\frac{1-\cos A}{2}}
  $

• Cosine Half-Angle Formula: $
  \cos \frac{A}{2}= \pm \sqrt{\frac{1+\cos A}{2}}
  $

• Tangent Half-Angle Formula: $
  \tan \frac{A}{2}= \pm \sqrt{\frac{1-\cos A}{1+\cos A}}
  $ (or) $
  \tan \frac{A}{2}=\frac{\sin A}{1+\cos A}
  $ (or) $
  \tan \frac{A}{2}=\frac{1-\cos A}{\sin A}
  $

Half Angle Formulas Derivation Using Double Angle Formulas

We know that the double angle formulas of sin ,cos, and tan are

• $
  \begin{gathered}
  \sin 2 x=2 \sin x \cos x
  \end{gathered}
  $

• $\cos 2 x=\cos^2 x-\sin^2 x$
  $\quad=1-2 \sin ^2 x$
  $\quad=2 \cos ^2 x-1$

• $
  \begin{aligned}
  & \tan 2 x=\frac{2 \tan x}{1-\tan ^2 x}
  \end{aligned}
  $

  
  

If we replace $x$ with $A / 2$ on both sides of every equation of the double-angle formulas, we get the halfangle identities (as $2 x=2(A / 2)=A$ ).
- $\sin A=2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)$
- $\cos A=\cos ^2\left(\frac{A}{2}\right)-\sin ^2\left(\frac{A}{2}\right)$

$
\quad =1-2 \sin ^2\left(\frac{A}{2}\right)
$

$\quad=2 \cos ^2\left(\frac{A}{2}\right)-1$

- $\tan A=\frac{2 \tan \left(\frac{A}{2}\right)}{1-\tan ^2\left(\frac{A}{2}\right)}$

Half-Angle Formula for Sine

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

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

Derivation of Half Angle Formula of sine

$
\cos \mathrm{A}=1-2 \sin ^2 \frac{\mathrm{A}}{2} \Rightarrow \sin ^2 \frac{\mathrm{A}}{2}=\frac{1-\cos \mathrm{A}}{2}
$

Now, for any $\triangle A B C$,

$
\cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}}
$

Using the above two formulas

$
\begin{aligned}
\sin ^2 \frac{A}{2} & =\frac{1}{2}\left[1-\frac{b^2+c^2-a^2}{2 b c}\right] \\
& =\frac{1}{2}\left[\frac{2 \mathrm{bc}-\mathrm{b}^2-\mathrm{c}^2+\mathrm{a}^2}{2 \mathrm{bc}}\right] \\
& =\frac{1}{2}\left[\frac{\mathrm{a}^2-(\mathrm{b}-\mathrm{c})^2}{2 \mathrm{bc}}\right] \\
& =\frac{(\mathrm{a}-\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c})}{4 \mathrm{bc}} \\
& =\frac{(2 \mathrm{~s}-2 \mathrm{~b})(2 \mathrm{~s}-2 \mathrm{c})}{4 \mathrm{bc}} \quad[\because \mathrm{a}+\mathrm{b}+\mathrm{c}=2 \mathrm{~s}] \\
\Rightarrow \sin ^2 \frac{\mathrm{A}}{2} & =\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{bc}}
\end{aligned}
$

As $0<\frac{\mathrm{A}}{2}<\frac{\pi}{2}$, so $\sin \frac{\mathrm{A}}{2}>0$

$
\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}
$

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

$\begin{aligned}
\cos \frac{A}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}} \\
\cos \frac{\mathrm{B}}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{b})}{\mathrm{ac}}} \\
\cos \frac{\mathrm{C}}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}}
\end{aligned}$

Derivation of Half Angle Formula of Cosine
We know that

$
\cos \mathrm{A}=2 \cos ^2 \frac{\mathrm{A}}{2}-1 \Rightarrow \cos ^2 \frac{\mathrm{A}}{2}=\frac{1+\cos \mathrm{A}}{2}
$

And for any $\triangle A B C, \quad \cos A=\frac{b^2+c^2-a^2}{2 b c}$
Using the above two formulas

$
\begin{aligned}
\cos ^2 \frac{A}{2} & =\frac{1}{2}\left[1+\frac{b^2+c^2-a^2}{2 b c}\right] \\
& =\frac{1}{2}\left[\frac{2 b c+b^2+c^2-a^2}{2 b c}\right] \\
& =\frac{1}{2}\left[\frac{(b+c)^2-a^2}{2 b c}\right] \\
& =\frac{(b+c+a)(b+c-a)}{4 b c} \\
& =\frac{(b+c+a)(a+b+c-2 a)}{4 b c}
\end{aligned}
$
Similarly, we can derive other formulas

Half Angle Formula for tan

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

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

This half-angle formula can be proved using $\tan x=\sin x / \cos x$, 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 $\pm$ 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 $\alpha / 2$ lies.

1. $\sin \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1-\cos \alpha}{2}}$
2. $\cos \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1+\cos \alpha}{2}}$
3. $\tan \left(\frac{\alpha}{2}\right)= \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}$

Derivation of Half Angle Formula of sine

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

$
\begin{aligned}
\sin ^2 \theta & =\frac{1-\cos (2 \theta)}{2} \\
\sin ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \left(2 \cdot \frac{\alpha}{2}\right)}{2} \\
& =\frac{1-\cos \alpha}{2} \\
\sin \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1-\cos \alpha}{2}}
\end{aligned}
$

Derivation of Half Angle Formula of Cosine

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

$
\begin{aligned}
\cos ^2 \theta & =\frac{1+\cos (2 \theta)}{2} \\
\cos ^2\left(\frac{\alpha}{2}\right) & =\frac{1+\cos \left(2 \cdot \frac{\alpha}{2}\right)}{2} \\
& =\frac{1+\cos \alpha}{2}
\end{aligned}
$
Derivation of Half Angle Formula of Tangent

$
\begin{aligned}
\tan ^2 \theta & =\frac{1-\cos (2 \theta)}{1+\cos (2 \theta)} \\
\tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \left(2 \cdot \frac{\alpha}{2}\right)}{1+\cos \left(2 \cdot \frac{\alpha}{2}\right)} \\
& =\frac{1-\cos \alpha}{1+\cos \alpha} \\
\tan \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}
\end{aligned}
$

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.

We know that the semi-perimeter of the triangle is

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

From this, we have

$
2 s=a+b+c
$

Derivation of Half-Angle Formula for Cosine

$
\begin{gathered}
\cos A=2 \cos ^2\left(\frac{A}{2}\right)-1 \quad \text { (or) } \\
2 \cos ^2\left(\frac{A}{2}\right)=1+\cos A
\end{gathered}
$

Now using the law of cosines,

$
\begin{gathered}
2 \cos ^2\left(\frac{A}{2}\right)=1+\frac{b^2+c^2-a^2}{2 b c} \\
2 \cos ^2\left(\frac{A}{2}\right)=\frac{2 b c+b^2+c^2-a^2}{2 b c} \\
2 \cos ^2\left(\frac{A}{2}\right)=\frac{(b+c)^2-a^2}{2 b c} \\
2 \cos ^2\left(\frac{A}{2}\right)=\frac{(b+c+a)(b+c-a)}{2 b c} \\
2 \cos ^2\left(\frac{A}{2}\right)=\frac{2 s(2 s-2 a)}{2 b c} \\
2 \cos ^2\left(\frac{A}{2}\right)=\frac{2 s(s-a)}{b c} \\
\cos ^2\left(\frac{A}{2}\right)=\frac{s(s-a)}{b c} \\
\cos ^2\left(\frac{A}{2}\right)=\sqrt{\frac{s(s-a)}{b c}}
\end{gathered}
$

Similarly, we can derive other half-angle identities for cosine using the semi-perimeter.

Derivation of Half-Angle Formula for Sine

$
\sin ^2\left(\frac{A}{2}\right)=\frac{1-\cos A}{2}
$

Using the law of cosines:

$
\begin{gathered}
\sin ^2\left(\frac{A}{2}\right)=\frac{1}{2}\left[1-\frac{b^2+c^2-a^2}{2 b c}\right] \\
=\frac{1}{2} \cdot \frac{a^2-(b-c)^2}{2 b c} \\
=\frac{1}{2} \cdot \frac{(a+b-c)(a+c-b)}{2 b c} \\
=\frac{1}{2} \cdot \frac{(a+b+c-2 c)(a+b+c-2 b)}{2 b c} \\
=\frac{1}{2} \cdot \frac{(2 s-2 c)(2 s-2 b)}{2 b c} \\
=\frac{(s-b)(s-c)}{b c} \\
\Rightarrow \sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}
\end{gathered}
$

Half-Angle Formula for Tangent

The half-angle formula for the tangent function can be derived using:

$
\tan \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right)}
$

Recommended Video Based on Half-Angle Formula

Solved Examples Based on Half-Angle Formula

Example 1: In triangle $A B C$ if $\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$ are in H.P. then the sides of triangle $A B C$ will be in.

Solution

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

$
\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}
$

Hence, the answer is sides of triangle $A B C$ will be in $A P$

Example 2: If in a triangle $\mathrm{ABC},(s-a)(s-b)=s(s-c)$ then angle C is equal to.

$
\begin{aligned}
& \tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2} \text { are in H.P. } \rightarrow \frac{2}{\tan \frac{B}{2}}=\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{C}{2}} \\
& 2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \\
& 2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-a)^2+s(s-b)^2}{(s-a)(s-b)(s-c)}} \\
& 2 \sqrt{s(s-b)^2}=\sqrt{s\left[(s-a)^2+(s-b)^2\right]} \\
& 2(s-b)=(s-a)+(s-c) \\
& a+c=2 b \\
& a, b, c \text { is in } A . P \text {. }
\end{aligned}
$
Solution: Half-Angle Formula (in terms of perimeter and sides of the triangle.

$
\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}
$

$
\begin{aligned}
\tan \frac{C}{2} & =\sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \because(s-a)(s-b)=s(s-c) \\
\tan \frac{C}{2}=1 & =\tan \frac{\pi}{4} \\
C & =\frac{\pi}{2}
\end{aligned}
$

Hence, the answer is angle C is $\frac{\pi}{2}$

Example 3: The value of $\tan \frac{B}{2} \tan \frac{C}{2}$ if $\mathrm{b}+\mathrm{c}=2 \mathrm{a}$, is.
Solution

$
\begin{aligned}
& \tan \frac{B}{2} \tan \frac{C}{2} \\
& =\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\
& =\frac{s-a}{s} \\
& =\frac{b+c-a}{a+b+c} \\
& =\frac{1}{3}
\end{aligned}
$

Hence, the answer is $\frac{1}{3}$

Example 4: In triangle ABC if $\tan \frac{A}{2}=\frac{3}{8}$ and $\tan \frac{C}{2}=\frac{8}{9}$ then sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in
Solution

$
\begin{aligned}
& \tan \frac{A}{2} \tan \frac{C}{2} \\
& =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\
& =\frac{s-b}{b} \\
& =\frac{a+c-b}{a+b+c}=\frac{3}{9}
\end{aligned}
$

(since $\tan \frac{A}{2}=\frac{3}{8}$ and $\tan \frac{C}{2}=\frac{8}{9}$ )

$
\begin{aligned}
& \Rightarrow 3(a+c-b)=a+b+c \\
& a+c=2 b
\end{aligned}
$

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

Example 5: In a $\triangle A B C \cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{\mathrm{C}}{2}$ is equal to

Solution: Trigonometric Ratios of Functions -
$\operatorname{cosec} \theta=\frac{H y p}{O p p}$
$\sec \theta=\frac{H y p}{\text { Base }}$
$\cot \theta=\frac{\text { Base }}{\text { Opp }}$

- wherein
$
\begin{aligned}
& \cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2} \\
& =\frac{s(s-a)}{\Delta} \cdot \frac{s(s-b)}{\Delta} \cdot \frac{s(s-c)}{\Delta} \\
& =\frac{s^3(a+b+c)}{\Delta^3}=\frac{s^2 \Delta^2}{\Delta^3}=\frac{s^2}{\Delta} \\
& =\left(\frac{\Delta^2}{r^2}\right) \times \frac{1}{\Delta}=\frac{\Delta}{r^2} \\
& \cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}=\frac{\Delta}{r^2}
\end{aligned}
$

Hence, the answer is $\frac{\Delta}{r^2}$