Why do we need half angle formulas?

Half angle formulas are important because they allow us to solve problems involving half angles without directly measuring them. They're useful in various applications, including engineering, physics, and advanced mathematics, where we often encounter situations that require working with half angles.

What's the connection between half angle formulas and double angle formulas?

Half angle formulas and double angle formulas are closely related. In fact, half angle formulas can be derived from double angle formulas. They're essentially inverse operations of each other. While double angle formulas express trigonometric functions of 2θ in terms of θ, half angle formulas express trigonometric functions of θ/2 in terms of θ.

Can half angle formulas be used to derive other trigonometric identities?

Yes, half angle formulas can be used to derive other trigonometric identities. For example, by using the half angle formulas for sine and cosine, you can derive the formula for sin(A+B). Half angle formulas are also used in deriving some power reduction formulas and other advanced trigonometric identities.

How do half angle formulas relate to the concept of angle bisectors in geometry?

Half angle formulas in trigonometry are closely related to angle bisectors in geometry. Just as an angle bisector divides an angle into two equal parts, half angle formulas allow us to work with half of a given angle. Geometrically, the point where an angle bisector intersects the unit circle has coordinates that can be expressed using half angle formulas.

How is the half angle formula for sine derived?

The half angle formula for sine is derived using the cosine double angle formula and the Pythagorean identity. We start with cos(2θ) = 1 - 2sin²θ, substitute θ with x/2, and then solve for sin(x/2). This process leads to the formula: sin(x/2) = ±√((1-cos(x))/2).

What is the half angle formula in trigonometry?

The half angle formula in trigonometry is a set of equations that allow us to express trigonometric functions of half an angle in terms of trigonometric functions of the full angle. These formulas are particularly useful when we need to find the sine, cosine, or tangent of half an angle without using a calculator.

How do half angle formulas relate to the concept of function composition?

Half angle formulas are examples of function composition in trigonometry. They express trigonometric functions of θ/2 in terms of trigonometric functions of θ. This is a form of composite function where we're essentially applying a trigonometric function to half of the input of another trigonometric function.

Can half angle formulas be used for any angle?

Yes, half angle formulas can be used for any angle. However, it's important to remember that when dealing with angles greater than 360° or less than 0°, you should first reduce the angle to its equivalent angle between 0° and 360° before applying the formula.

How do half angle formulas relate to the unit circle?

Half angle formulas can be visualized on the unit circle. For example, the sine half angle formula represents the y-coordinate of the point where the angle bisector of a central angle intersects the circle. Similarly, the cosine half angle formula represents the x-coordinate of this point.

Can half angle formulas be used in trigonometric form of complex numbers?

Yes, half angle formulas can be applied when working with the trigonometric form of complex numbers. When a complex number is expressed as r(cos(θ) + i sin(θ)), half angle formulas can be used to find the square root of this number or to halve the argument. This application is particularly useful in solving certain types of complex equations and in understanding the geometry of complex numbers.

What's the half angle formula for cosine?

The half angle formula for cosine is: cos(θ/2) = ±√((1+cos(θ))/2). Like the sine formula, it has a ± sign because the square root can be positive or negative. The choice of sign depends on the quadrant of θ/2.

Can you explain the ± sign in the half angle formula for sine?

The ± sign in the sine half angle formula (sin(x/2) = ±√((1-cos(x))/2)) indicates that there are two possible solutions. This is because the square root can be positive or negative. The correct sign depends on which quadrant the half angle (x/2) is in. It's positive in the first and second quadrants, and negative in the third and fourth quadrants.

How do you determine which sign to use in the half angle formula?

To determine which sign to use in the half angle formula, you need to consider the quadrant in which the half angle (θ/2) lies. For sine and cosine, use the positive sign if θ/2 is in the first or second quadrant, and the negative sign if it's in the third or fourth quadrant. For tangent, use the sign that makes the numerator and denominator have the same sign as tan(θ/2).

How is the half angle formula for tangent different from sine and cosine?

The half angle formula for tangent is unique because it doesn't involve a square root. It's expressed as: tan(θ/2) = ±√((1-cos(θ))/(1+cos(θ))) = sin(θ)/(1+cos(θ)) = (1-cos(θ))/sin(θ). This formula provides a direct way to calculate the tangent of a half angle without needing to calculate sine and cosine separately.

Why does the tangent half angle formula have two forms?

The tangent half angle formula has two main forms: tan(θ/2) = ±√((1-cos(θ))/(1+cos(θ))) and tan(θ/2) = sin(θ)/(1+cos(θ)) = (1-cos(θ))/sin(θ). These are algebraically equivalent but can be more or less useful in different situations. The first form shows the relationship to the cosine of the full angle, while the latter forms are often more practical for calculations.

How do half angle formulas relate to the concept of trigonometric substitution in calculus?

Half angle formulas are often used in trigonometric substitution, a technique in calculus for evaluating certain integrals. For example, when integrating √(1-x²), we might substitute x = sin(θ), which leads to √(1-x²) = cos(θ). The half angle formulas can then be used to express the result in terms of the original variable x. This connection demonstrates the broader applicability of half angle formulas beyond basic trigonometry.

What's the geometric interpretation of the half angle formulas?

Geometrically, half angle formulas can be interpreted as finding the coordinates of the point where the angle bisector of a central angle intersects the unit circle. The sine half angle formula gives the y-coordinate, the cosine half angle formula gives the x-coordinate, and the tangent half angle formula represents the slope of this angle bisector.

How do you apply half angle formulas in solving trigonometric equations?

Half angle formulas are useful in solving trigonometric equations, especially when the equation involves half angles. You can use these formulas to express the half angles in terms of full angles, making the equation easier to solve. For example, if you have an equation with sin(θ/2), you can replace it with ±√((1-cos(θ))/2) and then solve for θ.

Why do we need different formulas for sine, cosine, and tangent half angles?

We need different formulas for sine, cosine, and tangent half angles because these trigonometric functions represent different aspects of an angle in relation to the unit circle. Sine relates to the y-coordinate, cosine to the x-coordinate, and tangent to the slope. Each formula is derived differently and provides unique information about the half angle.

How do you choose which half angle formula to use in a problem?

The choice of which half angle formula to use depends on the information given in the problem and what you need to find. If you're given cos(θ) and need to find sin(θ/2), you'd use the sine half angle formula. If you're given sin(θ) and cos(θ) and need to find tan(θ/2), the tangent half angle formula would be most useful. Always consider what information you have and what you need to calculate.

Can half angle formulas be used in reverse to find full angles?

Yes, half angle formulas can be used in reverse to find full angles from half angles. For example, if you know sin(θ/2), you can use the relationship sin²(θ/2) = (1-cos(θ))/2 to find cos(θ), and from there determine θ. However, this process often involves solving quadratic equations and considering multiple possible solutions.

Why do some half angle formulas have square roots while others don't?

The sine and cosine half angle formulas involve square roots because they're derived from quadratic equations. These square roots represent the two possible solutions for the half angle. The tangent half angle formula, however, can be simplified to a form without square roots because the square roots in the numerator and denominator cancel out.

How do half angle formulas change when dealing with radians instead of degrees?

The half angle formulas remain the same whether you're working with degrees or radians. The formulas are based on the relationships between trigonometric functions, which are independent of the angle measurement system. However, when applying the formulas, you need to ensure that you're consistent in using either degrees or radians throughout your calculations.

Can half angle formulas be extended to third angle or quarter angle formulas?

While there are no standard "third angle" or "quarter angle" formulas, the concept can be extended. For example, you could use the half angle formula twice to get a quarter angle formula. However, these quickly become complex and are rarely used in practice. For smaller fractional angles, it's often more practical to use other methods like the angle addition formulas.

How do half angle formulas relate to the concept of even and odd functions?

Half angle formulas highlight the even/odd properties of trigonometric functions. For example, the cosine half angle formula cos(θ/2) = ±√((1+cos(θ))/2) shows that cosine is an even function (cos(-θ) = cos(θ)), as changing the sign of θ doesn't change the result. The sine and tangent formulas, on the other hand, reflect their odd function nature.

Why is the tangent half angle formula often considered the most useful?

The tangent half angle formula is often considered the most useful because it doesn't involve square roots in its simplified form (tan(θ/2) = sin(θ)/(1+cos(θ)) = (1-cos(θ))/sin(θ)). This makes it easier to use in many calculations and algebraic manipulations. Additionally, it directly relates the half angle to both sine and cosine of the full angle, providing a versatile tool for problem-solving.

How do half angle formulas relate to the concept of inverse trigonometric functions?

Half angle formulas can be used in conjunction with inverse trigonometric functions. For example, if you know cos(θ) and want to find θ/2, you could use the cosine half angle formula and then apply the arccosine function. This relationship is particularly useful in solving equations involving inverse trigonometric functions.

How do half angle formulas relate to the concept of limits in calculus?

Half angle formulas are related to limits in calculus, particularly in the context of small angle approximations. As θ approaches 0, sin(θ/2) ≈ θ/2, cos(θ/2) ≈ 1, and tan(θ/2) ≈ θ/2. These approximations, derived from half angle formulas, are useful in many calculus applications, including the derivation of derivative formulas for trigonometric functions.

Why is it important to consider the domain when using half angle formulas?

Considering the domain is crucial when using half angle formulas because the formulas involve square roots and fractions. For example, the expression under the square root in the sine and cosine formulas must be non-negative. Similarly, the denominator in the tangent formula must not be zero. Understanding these domain restrictions helps avoid errors and correctly interpret results.

Half Angle Formula

Maths
Trigonometric Functions
Half Angle Formula

Half Angle Formula

Komal MiglaniUpdated on 02 Jul 2025, 07:31 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.

This Story also Contains

What are Half Angle Formula ?
Derivation of Half Angle Formulas
Solved Example Based on Half Angle Formula

Half Angle Formula

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.

What are Half Angle Formula ?

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

Note that 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.

Derivation of Half Angle Formulas

The half-angle formula can be derived with the help of the reduction formula.
Reduction formulas are:
$\sin ^2 \theta=\frac{1-\cos (2 \theta)}{2}$
$\cos ^2 \theta=\frac{1+\cos (2 \theta)}{2}$
$\tan ^2 \theta=\frac{1-\cos (2 \theta)}{1+\cos (2 \theta)}$

Derivation of Half Angle Formula for 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 for 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} \\
\cos \left(\frac{\alpha}{2}\right) & = \pm \sqrt{\frac{1+\cos \alpha}{2}}
\end{aligned}
$

Derivation of Half Angle Formula for Tangent

For the tangent identity, we have

$
\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}
$

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

$
\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}
$
We know that,

$
\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 $\triangle A B C, \cos A=\frac{b^2+c^2-a^2}{2 b c}$

Using the above two formulas

In a similar way, we can derive other 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 b c-b^2-c^2+a^2}{2 b c}\right] \\
& =\frac{1}{2}\left[\frac{a^2-(b-c)^2}{2 b c}\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}} \\
& \text { As } 0<\frac{\mathrm{A}}{2}<\frac{\pi}{2} \text {, so } \sin \frac{\mathrm{A}}{2}>0 \\
& \sin \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{bc}}}
\end{aligned}
$

Derivation of Half-Angle Formula for Cosine

$
\begin{aligned}
& \cos \frac{\mathrm{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}
$
We know that

$
\begin{aligned}
& \cos \mathrm{A}=2 \cos ^2 \frac{\mathrm{A}}{2}-1 \Rightarrow \cos ^2 \frac{\mathrm{A}}{2}=\frac{1+\cos \mathrm{A}}{2} \\
& \text { And for any } \triangle \mathrm{ABC}, \cos \mathrm{A}=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2 \mathrm{bc}}
\end{aligned}
$

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

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

$
\cos \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{a})}{\mathrm{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.

$
\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 $\tan x= \frac{\sin x}{\cos x}$ and using the half-angle formula of sine and cosine.

Solved Example Based on Half Angle Formula

Example 1: If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), \mathrm{y}, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression, then $|x-2 y|$ is equal to [JEE MAINS 2021]
Solution

$
\begin{aligned}
& x=\frac{\tan \left(\frac{\pi}{9}\right)+\tan \left(\frac{7 \pi}{18}\right)}{2} \text { and } \\
& y=\frac{\tan \left(\frac{\pi}{9}\right)+\tan \left(\frac{5 \pi}{18}\right)}{2}
\end{aligned}
$
$
\begin{aligned}
& \therefore x-2 y=\frac{1}{2}\left[\tan \left(\frac{\pi}{9}\right)+\tan \left(\frac{7 \pi}{18}\right)-2 \tan \left(\frac{\pi}{9}\right)-2 \tan \left(\frac{5 \pi}{18}\right)\right] \\
& =\frac{1}{2}\left[\tan \left(\frac{7 \pi}{18}\right)-\tan \left(\frac{\pi}{9}\right)-2 \tan \left(\frac{5 \pi}{18}\right)\right] \\
& =\frac{1}{2}\left[\tan \left(\frac{\pi}{2}-\frac{\pi}{9}\right)-\tan \left(\frac{\pi}{9}\right)-2 \tan \left(\frac{5 \pi}{18}\right)\right] \\
& =\frac{1}{2}\left[\cot \left(\frac{\pi}{9}\right)-\tan \left(\frac{\pi}{9}\right)-2 \tan \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)\right] \\
& =\frac{1}{2}\left[\frac{\cos \left(\frac{\pi}{9}\right)}{\sin \left(\frac{\pi}{9}\right)}-\frac{\sin \left(\frac{\pi}{9}\right)}{\cos \left(\frac{\pi}{9}\right)}-2 \cot \left(\frac{2 \pi}{9}\right)\right] \\
& =\frac{1}{2}\left[\frac{2 \cos (2 \pi / 9)}{\sin \left(\frac{2 \pi}{9}\right)}-2 \cot \left(\frac{2 \pi}{9}\right)\right] \\
& =0
\end{aligned}
$

Hence, the answer is 0

Example 2: The value of $2 \sin \left(\frac{\pi}{8}\right) \sin \left(\frac{2 \pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) \sin \left(\frac{5 \pi}{8}\right) \sin \left(\frac{6 \pi}{8}\right) \sin \left(\frac{7 \pi}{8}\right)$ is :
[JEE MAINS 2021]
Solution:

$
\begin{aligned}
& \text { As } \sin \left(\frac{5 \pi}{8}\right)=\sin \left(\pi-\frac{3 \pi}{8}\right)=\sin \left(\frac{3 \pi}{8}\right) \\
& \text { similarly } \sin \left(\frac{6 \pi}{8}\right)=\sin \left(\frac{2 \pi}{8}\right) \text { and } \sin \left(\frac{7 \pi}{8}\right)=\sin \left(\frac{\pi}{8}\right)
\end{aligned}
$

$\therefore$ Required value

$
\begin{aligned}
& =2 \cdot \sin ^2\left(\frac{\pi}{8}\right) \cdot \sin ^2\left(\frac{2 \pi}{8}\right) \cdot \sin ^2\left(\frac{3 \pi}{8}\right) \\
& =2 \cdot\left(\frac{1}{\sqrt{2}}\right)^2 \cdot \sin ^2\left(\frac{\pi}{8}\right) \cdot \sin ^2\left(\frac{3 \pi}{8}\right) \\
& =\sin ^2\left(\frac{\pi}{8}\right) \cdot \cos ^2\left(\frac{\pi}{8}\right) \\
& =\frac{4}{4} \sin ^2\left(\frac{\pi}{8}\right) \cdot \cos ^2\left(\frac{\pi}{8}\right) \\
& =\frac{1}{4}\left(\sin \left(\frac{\pi}{4}\right)\right)^2=\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8}
\end{aligned}
$
Hence, the answer is $1 / 8$

Example 3: If $\cos \alpha+\cos \beta=\frac{3}{2}$ and $\sin \alpha+\sin \beta=\frac{1}{2}$ and $\theta \quad$ is the arithmetic mean of $\alpha$ and $\beta$, then $\sin 2 \theta+\cos 2 \theta$ is equal to : [JEE MAINS 2015]
Solution:

Trigonometric Ratios of Submultiples of an Angle -

$
\begin{gathered}
\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\
\sin C+\sin D=2 \cdot \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right) \\
\cos C+\cos D=2 \cdot \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)
\end{gathered}
$
This shows the transformation formulae and double angle formulae.
Therefore,

$
\begin{aligned}
& \cos \alpha+\cos \beta=\frac{3}{2} \Rightarrow 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{2} \\
& \sin \alpha+\sin \beta=\frac{1}{2} \Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{1}{2}
\end{aligned}
$

dividing (2) by (1), we get

$
\tan \frac{(\alpha+\beta)}{2}=\frac{1}{3} \Rightarrow \tan \theta=\frac{1}{3}
$

$
\begin{aligned}
& \Rightarrow \tan 2 \theta=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4} \\
& \Rightarrow \sin 2 \theta=\frac{3}{5}, \quad \cos 2 \theta=\frac{4}{5}
\end{aligned}
$

[By making a triangle]
Thus,

$
\Rightarrow \sin 2 \theta+\cos 2 \theta=\frac{7}{5}
$
Hence, the answer is $7 / 5$

Example 4: Let $\alpha, \beta$ be such that $\pi<\alpha-\beta<3 \pi$. If $\sin \alpha+\sin \beta=-21 / 65$, and
$\cos \alpha+\cos \beta=-27 / 65$ then the value of $\cos \frac{\alpha-\beta}{2}$ is :

Solution:

$\sin \alpha+\sin \beta=\frac{-21}{65} \Rightarrow 2 \sin \frac{(\alpha+\beta)}{2} \cos \frac{(\alpha-\beta)}{2}=\frac{-21}{65}$ $\qquad$

$
\cos \alpha+\cos \beta=\frac{-27}{65} \Rightarrow 2 \cos \frac{(\alpha+\beta)}{2} \cos \frac{(\alpha-\beta)}{2}=\frac{-21}{65}
$
Squaring and adding,

$
\begin{aligned}
& 4 \cos ^2 \frac{(\alpha-\beta)}{2}\left[\sin ^2 \frac{\alpha+\beta}{2}+\cos ^2 \frac{\alpha+\beta}{2}\right]=\frac{21^2+27^2}{65^2} \\
& \Rightarrow \cos ^2 \frac{(\alpha-\beta)}{2}=\frac{1170}{4 \times 4225} \Rightarrow \cos ^2 \frac{(\alpha-\beta)}{2}=\frac{1170}{4 \times 4225}=\frac{9}{130} \\
& \Rightarrow \cos \frac{(\alpha-\beta)}{2}=\frac{-3}{\sqrt{130}} \\
& {\left[\because \pi<\alpha-\beta<3 \pi \Rightarrow \frac{\pi}{2} \leq \frac{\alpha-\beta}{2} \leq \frac{3 \pi}{2} \Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)<0\right]} \\
& \text { Hence, the answer is }-\frac{3}{\sqrt{130}}
\end{aligned}
$

Example 5: Find the range of functions $\tan \frac{\alpha}{2} \cdot \tan \alpha$
Solution:

$
\begin{aligned}
& \quad \tan \frac{\alpha}{2} \cdot \tan \alpha=\left(\frac{1-\cos \alpha}{\sin \alpha}\right) \cdot \frac{\sin \alpha}{\cos \alpha} \\
& =\sec \alpha-1 \\
& \text { Range of } \sec \alpha(-\infty,-1] U [ 1, \infty) \\
& \text { Range of } f(\alpha) \text { is }(-\infty,-2] U [0, \infty)
\end{aligned}
$

Hence, the answer is $(-\infty,-2] U [0, \infty)$

Frequently Asked Questions (FAQs)

Q: How do half angle formulas relate to the concept of trigonometric identities?

Half angle formulas are themselves trigonometric identities, and they can be used to derive or verify other identities. For example, the Pythagorean identity sin²(θ/2) + cos²(θ

Q: Can half angle formulas be used in solving trigonometric inequalities?

Yes, half angle formulas can be useful in solving trigonometric inequalities, especially when the inequality involves half angles. By using these formulas, you can often transform the inequality into a more standard form involving full angles, which may be easier to solve. However, it's crucial to consider the domain restrictions and possible sign changes when applying these formulas in inequalities.

Q: Why is it important to understand the derivation of half angle formulas?

Understanding the derivation of half angle formulas is crucial because it provides insight into the relationships between trigonometric functions and helps in remembering and applying the formulas correctly. The derivation process also reinforces important trigonometric concepts and identities, enhancing overall understanding of trigonometry. Moreover, the techniques used in these derivations can be applied to solve other trigonometric problems.

Q: How do half angle formulas relate to the concept of rational trigonometric functions?

Half angle formulas play a role in creating rational trigonometric functions. For example, the tangent half angle formula tan(θ/2) = sin(θ)/(1+cos(θ)) expresses tan(θ/2) as a rational function of sin(θ) and cos(θ). This relationship is useful in various areas of mathematics, including integration techniques and solving certain types of differential equations.

Q: How do half angle formulas relate to the concept of periodicity in trigonometric functions?

Half angle formulas highlight the relationship between the periodicity of full angle and half angle functions. While sin(θ) and cos(θ) have a period of 2π, sin(θ/2) and cos(θ/2) have a period of 4π. This doubling of the period is a direct result of halving the angle in the argument of the function, and it's an important consideration when graphing or analyzing these functions.

Q: Why is it sometimes necessary to use multiple half angle formulas in a single problem?

In complex trigonometric problems, it's often necessary to use multiple half angle formulas because different parts of the problem may require different trigonometric ratios of the half angle. For example, you might need to find both sin(θ/2) and cos(θ/2) to solve a problem. Using multiple formulas allows you to approach the problem from different angles and verify your results.

Q: How do half angle formulas relate to the concept of symmetry in trigonometric functions?

Half angle formulas reflect the symmetry properties of trigonometric functions. For instance, the fact that cos(θ/2) = ±√((1+cos(θ))/2) shows the even symmetry of cosine (cos(-θ) = cos(θ)), while the form of the sine half angle formula reflects sine's odd symmetry (sin(-θ) = -sin(θ)). Understanding these symmetries can help in choosing the correct sign in half angle calculations.

Q: Can half angle formulas be used in solving systems of trigonometric equations?

Yes, half angle formulas can be very useful in solving systems of trigonometric equations, especially when the system involves both full angles and half angles. By using these formulas, you can often express all terms in the system using the same angle, making the system easier to solve. This technique is particularly helpful in more complex trigonometric problems and proofs.

Q: Can half angle formulas be used in complex number theory?

Yes, half angle formulas can be extended to complex number theory. In fact, they play a role in de Moivre's formula and Euler's formula, which are fundamental in relating complex exponentials to trigonometric functions. These extensions of half angle formulas are crucial in many areas of advanced mathematics and physics.

Q: How do half angle formulas relate to trigonometric substitution in integration?

Half angle formulas are often used in trigonometric substitution, a technique in integration. For instance, when integrating expressions involving √(1-x²), we might substitute x = sin(θ), which then leads to √(1-x²) = cos(θ). The half angle formulas can then be used to simplify the resulting trigonometric expressions, making the integration more manageable.

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