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Half Angle Formula - Practice Questions & MCQ

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

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

  • 11 Questions around this concept.

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QuestionMEDIUM

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 :

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

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

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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}

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Half Angle Formula

Study other Related Concepts

Half Angle Formula Current Topic

Measuring Angles
Trigonometric Ratios
Trigonometric Identities
Sign of Trigonometric Functions
Graphs of General Trigonometric Functions
Trigonometric Ratios of Allied Angles
Trigonometric Ratios of Compound Angles
Sum-to-Product and Product-to-Sum Formulas
Product To Sum Formulas
Double Angle Formulas

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Half Angle Formula

Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

Page No. : 3.11

Line : 34

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