Class 11: Trigonometric Formulas – Lecture Notes – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Cosine of the difference and sum of two numbers

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

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

Sine of the difference and sum of two numbers

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

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

Tangent of the difference and sum of two numbers

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

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

Few more derivations based on the above formulas

$\displaystyle \sin ( A + B ) \sin ( A - B ) = \sin^2 A - \sin ^2 B = \cos^2 B - \cos^2 A$

$\displaystyle \cos ( A + B ) \cos ( A - B ) = \cos^2 A - \sin ^2 B = \cos^2 B - \sin^2 A$

$\displaystyle \sin ( A + B + C) = \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C$

$\displaystyle \cos ( A + B + C) = \cos A \cos B \cos C - \cos A \sin B \sin C - \sin A \cos B \sin C - \sin A \sin B \cos C$

$\displaystyle \tan ( A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A }$

Formula to transform the product into sum or difference

$\displaystyle 2 \sin A \cos B = \sin ( A + B ) + \sin ( A - B)$

$\displaystyle 2 \cos A \sin B = \sin ( A + B ) - \sin ( A - B)$

$\displaystyle 2 \cos A \cos B = \cos( A + B ) + \cos ( A - B)$

$\displaystyle 2 \sin A \sin B = \cos ( A - B ) - \cos ( A + B)$

Formula to transform the sum or difference into product

$\displaystyle \sin C + \sin D = 2 \sin \Big( \frac{C+D}{2} \Big) \cos \Big( \frac{C-D}{2} \Big)$

$\displaystyle \sin C - \sin D = 2 \sin \Big( \frac{C-D}{2} \Big) \cos \Big( \frac{C+D}{2} \Big)$

$\displaystyle \cos C + \cos D = 2 \cos \Big( \frac{C+D}{2} \Big) \cos \Big( \frac{C-D}{2} \Big)$

$\displaystyle \cos D - \cos C = 2 \sin \Big( \frac{C+D}{2} \Big) \sin \Big( \frac{C-D}{2} \Big)$

$\displaystyle \cos C - \cos D = -2 \sin \Big( \frac{C+D}{2} \Big) \sin \Big( \frac{C-D}{2} \Big)$

Value of trigonometric functions at $\displaystyle 2x$ in terms of values at $\displaystyle x$

$\displaystyle \sin 2 x = 2 \sin x \cos x$

$\displaystyle \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\displaystyle \tan 2x = \frac{2 \tan x }{1- \tan^2 x}$

$\displaystyle \sin 2x = \frac{2 \tan x }{1+ \tan^2 x}$

$\displaystyle \cos 2x = \frac{1 - \tan^2 x }{1+ \tan^2 x}$

Value of trigonometric functions at $\displaystyle 3x$ in terms of values at $\displaystyle x$

$\displaystyle \sin 3x = 3 \sin x - 4 \sin^3 x$

$\displaystyle \cos 3x = 4 \cos^3 x - 3 \cos x$

$\displaystyle \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}$

Value of trigonometric functions at $\displaystyle x$ in terms of values at $\displaystyle \frac{x}{3}$

$\displaystyle \sin x = 3 \sin \frac{x}{3} - 4 \sin^3 \frac{x}{3}$

$\displaystyle \cos x = 4 \cos^3 \frac{x}{3} - 3 \cos \frac{x}{3}$

$\displaystyle \tan x = \frac{3 \tan \frac{x}{3} - \tan^3 \frac{x}{3}}{1 - 3 \tan^2 \frac{x}{3}}$

Values of trigonometrical functions at some important points

$\displaystyle \sin \frac{\pi}{10} = \frac{\sqrt{5}-1}{4}$

$\displaystyle \cos \frac{\pi}{10} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$

$\displaystyle \cos \frac{\pi}{5} = \frac{\sqrt{5}+1}{4}$

$\displaystyle \sin \frac{\pi}{5} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4}$

Law of Sines or Sine Rule

$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Law of Cosines

$\displaystyle \cos A = \frac{b^2 + c^2 - a^2}{2 bc}$ or $\displaystyle a^2 = b^2 + c^2 - 2bc \cos A$

$\displaystyle \cos B = \frac{c^2 + a^2 - b^2}{2 ca}$ or $\displaystyle b^2 = c^2 + a^2 - 2ca \cos B$

$\displaystyle \cos C = \frac{a^2 + b^2 - c^2}{2 ab}$ or $\displaystyle c^2 = a^2 + b^2 - 2ab \cos A$

Projection Formula

$\displaystyle a = b \cos C + c \cos B$

$\displaystyle b = c \cos A + a \cos C$

$\displaystyle c = a \cos B + b \cos A$

Napier’s Analogy (law of tangents). In any $\displaystyle \triangle ABC$ , we have

$\displaystyle \tan \Big( \frac{B-C}{2} \Big) = \Big( \frac{b-c}{b+c} \Big) \cot \frac{A}{2}$

$\displaystyle \tan \Big( \frac{A-B}{2} \Big) = \Big( \frac{a-b}{a+b} \Big) \cot \frac{C}{2}$

$\displaystyle \tan \Big( \frac{C-A}{2} \Big) = \Big( \frac{c-a}{c+a} \Big) \cot \frac{B}{2}$

Area of a Triangle

$\displaystyle \triangle = \frac{1}{2} bc \sin A = \frac{1}{2} ca \sin B = \frac{1}{2} ab \sin C$

Solving trigonometric equations

Step 1: Find a value of $\displaystyle x$ , preferably between $\displaystyle 0$ and $\displaystyle 2\pi$ or between $\displaystyle -\pi$ and $\displaystyle \pi$ satisfying the given equation and call it $\displaystyle \alpha$

Step 2:

If the equation is $\displaystyle \sin x = \sin \alpha$ , then $\displaystyle x = n\pi + ( -1)^n \alpha, n \in Z$ as a general solution
If the equation is $\displaystyle \cos x = \cos \alpha$ , then $\displaystyle x = 2n\pi \pm \alpha, n \in Z$ as a general solution
If the equation is $\displaystyle \tan x = \tan \alpha$ , then $\displaystyle x = n\pi + \alpha, n \in Z$ as a general solution