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Half Angle Formula - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Half Angle Formula is considered one of the most asked concept.

  • 16 Questions around this concept.

Solve by difficulty

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 :

The value of  \sin\ \frac{\pi }{6} using half angle identity.

The value of \tan\ 60^{\circ} is

$If \sin \frac{\theta }{2}= \frac{2}{5} where 0^{\circ}< \theta < 90^{\circ}, then \tan \theta =$

$If \sin \frac{A}{3}= \frac{1}{5} ; 0^{\circ}< A< 90^{\circ},then \tan A=$

$If \sin A=\frac{2}{3},then\left | \sin \frac{A}{2}-\cos \frac{A}{2} \right |=$

$If \sin A=\frac{3}{5},then \left | \sin \frac{A}{2}+\cos \frac{A}{2} \right |=$

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What is the value of $\cot \frac{A}{2}$ if $\sin A=\frac{3}{5}$ ?

Concepts Covered - 1

Half Angle Formula

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

These formulae can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle.

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}
$
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}
$

For the tangent identity, we have

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

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Half Angle Formula

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