 # Trigonometry Formulas For Class 11

Trigonometry is a branch of mathematics which studies the relationships that involve angles and lengths of triangles. It is very important to mathematics subject as an element of statistics, linear algebra and calculus. In addition to mathematics, it also contributes majorly to engineering, physics, astronomy and architectural design. Trigonometry Formulas for class 11 plays a crucial role in solving any problem related to this chapter.

CBSE Class 11 mathematics contains Trigonometric functions where you need to learn many formulae. This chapter lays the foundation to Inverse trigonometric functions in class 12. So, it is very important chapter where students have to concentrate more. Trigonometric is a subject where there are formulas for everything and without which you cannot solve a problem.  Students always feel confused between these formulas. This is the reason, we at Byju’s provide all the important formulas at a single page for easy reading and comparing. We believe, Trigonometry Class 11 formulas provided here will help students to learn them and can have a quick glance when needed.

 Trigonometry Class 11 Formulas $\sin (-\theta ) = -\sin \theta$ $\cos (-\theta ) = \cos \theta$ $\tan (-\theta ) = -\tan \theta$ $cosec (-\theta ) = -cosec \theta$ $\sec (-\theta ) = \sec \theta$ $\cot (-\theta ) = -\cot \theta$ Product to Sum Formulas $\sin \, x \,\ sin \, y = \frac{1}{2}\left [ \cos\left ( x – y \right ) -\cos \left ( x+y \right ) \right ]$ $\cos\, x \, \cos\, y = \frac{1}{2}\left [ \cos \left ( x – y \right ) + \cos \left ( x+y \right ) \right ]$ $\sin\, x \, \cos\, y = \frac{1}{2}\left [ \sin\left ( x + y \right ) + \sin \left ( x-y \right ) \right ]$ $\cos\, x \, \sin\, y = \frac{1}{2}\left [ \sin\left ( x + y \right ) – \sin\left ( x-y \right ) \right ]$ Sum to Product Formulas $\sin\, x + \sin \, y = 2\, \sin \left ( \frac{x+y}{2} \right ) \cos \left ( \frac{x-y}{2} \right )$ $\sin\, x -\sin\, y = 2\, \cos \left ( \frac{x+y}{2} \right ) \sin \left ( \frac{x-y}{2} \right )$ $\cos \, x + \cos \, y = 2 \, \cos \left ( \frac{x+y}{2} \right ) \cos\left ( \frac{x-y}{2} \right )$ $\cos\, x -\cos\, y = – 2 \, \sin \left ( \frac{x+y}{2} \right ) \sin \left ( \frac{x-y}{2} \right )$

Basic Formulas

$\sin (A+B) = \sin A \cos B + \cos A \sin B$

$\sin (A -B) = \sin A \cos B – \cos A \sin B$

$\cos (A + B) = \cos A \cos B – \sin A \sin B$

$\cos (A – B) = \cos A \cos B + \sin A \sin B$

$\tan (A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B}$

$\tan (A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$

$\cos (A + B) \cos (A – B) = \cos^{2}A – \sin^{2}B = \cos^{2}B – \sin^{2}A$

$\sin (A + B) \sin (A – B) = \sin^{2}A – \sin^{2}B = \cos^{2}B – \cos^{2}A$

$\sin 2A = 2 \sin A \cos A = \frac{2\tan A}{1+\tan^{2}A}$

$\cos 2A = \cos^{A} – \sin^{2}A = 1 – 2sin^{2}A = 2cos^{2}A – 1 = \frac{1-\tan^{2}A}{1 + \tan^{2}A}$

$\tan 2A =\frac{2 \tan A}{1 – \tan^{2}A}$

$\sin 3A = 3\sin A – 4\sin^{3}A = 4\sin(60^{\circ}-A).\sin A .\sin( 60^{\circ}+A)$

$\cos 3A = 4\cos^{3}A – 3\cos A = 4\cos\left ( 60^{\circ}-A \right ).\cos A . \cos\left ( 60^{\circ} +A\right )$

$\tan 3A = \frac{3\tan A – \tan^{3}A}{1-3\tan^{2}A} = \tan\left ( 60^{\circ}-A \right ).\tan A . \tan\left ( 60^{\circ}+A\right )$

$\sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$

Example: Find the maximum value of $\cos^{2}\cos\theta + \sin^{2} \sin\theta$ for any real value of $\theta$

The maximum value of $\cos^{2}\cos\theta = 1$ and $\sin^{2}\sin\theta$ is $\sin^{2} 1$ where both exist for $\theta = \frac{\pi}{2}$. Therefore maximum value will be $1+ \sin^{2}1$..

1. chirasmita nanda

helped me a lot. thank you byju’s 🙂

2. Devanshu