Cosine of the difference and sum of two numbers

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

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

Sine of the difference and sum of two numbers

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

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

Tangent of the difference and sum of two numbers

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

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

Few more derivations based on the above formulas

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

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

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

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

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

Formula to transform the product into sum or difference

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

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

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

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

Formula to transform the sum or difference into product

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

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

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

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

or

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Values of trigonometrical functions at some important points

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

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

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

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

Law of Sines or Sine Rule

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

Law of Cosines

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

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

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

Projection Formula

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

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

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

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

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

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

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

Area of a Triangle

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

Solving trigonometric equations

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

Step 2:

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