Careers360 Logo
JEE Main Eligibility Criteria 2025- Marks in 12th, Age Limit, Number of Attempts

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.

  • 11 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

Concepts Covered - 1

Half Angle Formula

Half Angle Formula

\begin{aligned}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}}\end{aligned}

 

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.

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.

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}

Study it with Videos

Half Angle Formula

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Half Angle Formula

Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

Page No. : 3.11

Line : 34

E-books & Sample Papers

Get Answer to all your questions

Back to top