## TRIGONOMETRY FORMULAE

Trigonometry Formulae: Don’t try to remember just know the concept of derivation

With the help of sin (A + B), we can derive all the remaining trigonometric formulae. Let’s see, how …..

Compound Angle

(1) $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

(2) $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$

(3) $\cos \left( A+B \right)=\sin \left[ \frac{\pi }{2}-\left( A+B \right) \right]$

$=\sin \left[ \left( {}^{\pi }\!\!\diagup\!\!{}_{2}\;-A \right)-\left( B \right) \right]$

$=\sin \left( \frac{\pi }{2}-A \right)\cos B-\cos \left( \frac{\pi }{2}-A \right)\sin B$

$\therefore \,\,\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$

(4) $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$

(5) $\tan \left( A+B \right)=\frac{\sin \left( A+B \right)}{\cos \left( A+B \right)}$

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

Dividing Numerator and Denominator by

$\therefore \,\tan \left( A+B \right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

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

Similarly,

(7) $\cot \left( A+B \right)=\frac{\cot A\cot B-1}{\cot B+\cot A}$

(8) $\cot \left( A-B \right)=\frac{\cot A\cot B+1}{\cot B-\cot A}$

Transformation Formulae:

By adding (1) and (2), we get

(9) $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$

By subtracting (2) from (1), we get

(10) $2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)$

By adding (3) and (4), we get

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

By subtracting (4) from (3), we get

(12) $2\sin A\sin B=-\cos \left( A+B \right)+\cos \left( A-B \right)$

Now, consider $A+B=C$ and $A-B=D$

Then, $A=\frac{C+D}{2}\,$ and $B=\frac{C-D}{2}$

(13) From (9), $\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$

(14) From (10), $\sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}$

(15) From (11), $\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$

(16) From (12), $\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}$

Multiply (1) and (2), we get

(17) ${{\sin }^{2}}A-{{\sin }^{2}}B=\sin \left( A+B \right).\sin \left( A-B \right)$

${{\cos }^{2}}B-{{\cos }^{2}}A=\sin \left( A+B \right).\sin \left( A-B \right)$

(18) ${{\cos }^{2}}B-{{\sin }^{2}}A=\cos \left( A+B \right).\cos \left( A-B \right)$

${{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right).\cos \left( A-B \right)$

Multiple Angles

(19) $\sin \left( 2A \right)=\sin \left( A+A \right)$

$=\sin A\cos A+\cos A\sin A\,\,\,\,\,\,\,\,\,\,\left\{ \text{from }\left( 1 \right) \right\}$

$\therefore \,\,\sin \left( 2A \right)=2\sin A\cos A$

(20) $\cos \left( 2A \right)=\cos \left( A+A \right)$

$=\cos A\cos A-\sin A\sin A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \text{from}\,\left( 3 \right) \right\}$

$\therefore \,\,\cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A$

$=1-2{{\sin }^{2}}A\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \text{from}\,\,\,{{\sin }^{2}}A+{{\cos }^{2}}A=1 \right\}$

$=2{{\cos }^{2}}A-1$

(21) $\tan \left( 2A \right)=\tan \left( A+A \right)$

$=\frac{\tan A+\tan A}{1-\tan A\tan A}$

$\therefore \tan \left( 2A \right)\,\,=\frac{2\tan A}{1-{{\tan }^{2}}A}$

$\sin \left( 2A \right)\,\,\,\,\,=\frac{2\tan A}{1+{{\tan }^{2}}A}$

$\cos \left( 2A \right)\,\,\,=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$

(22) $\sin \left( 3A \right)=\sin \left( 2A+A \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin 2A\cos A+\cos 2A\sin A$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A\cos A\cos A+\left( 1-2{{\sin }^{2}}A \right)\sin A$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A\left( 1-{{\sin }^{2}}A \right)+\left( 1-2{{\sin }^{2}}A \right)\sin A$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin A-2{{\sin }^{3}}A+\sin A-2{{\sin }^{3}}A$

$\therefore \sin \left( 3A \right)=3\sin A-4{{\sin }^{3}}A$

(23) $\cos \left( 3A \right)=\cos \left( 2A+A \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\cos 2A\cos A-\sin 2A\sin A$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2\sin A\cos A.\sin A$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( 2{{\cos }^{2}}A-1 \right)\cos A-2\cos A\left( 1-{{\cos }^{2}}A \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2{{\cos }^{3}}A-\cos A-2\cos A+2{{\cos }^{3}}A$

$\therefore \cos \left( 3A \right)\,\,=4{{\cos }^{3}}A-3\cos A$

(24) $\tan \left( 3A \right)=\tan \left( 2A+A \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\tan 2A+\tan A}{1-\tan 2A\tan A}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\frac{2\tan A}{1-{{\tan }^{2}}A}+\tan A}{1-\frac{2\tan A}{1-{{\tan }^{2}}A}.\tan A}$

$\therefore \tan \left( 3A \right)=\frac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$

From (20), $\cos 2A=1-2{{\sin }^{2}}A$

And $\cos 2A=2{{\cos }^{2}}A-1$

(25) $1-\cos 2A=2{{\sin }^{2}}A$

(26) $1+\cos 2A=2{{\cos }^{2}}A$

(27) $1-\sin 2A={{\sin }^{2}}A+{{\cos }^{2}}A-2\sin A\cos A$

(28) $1+\sin 2A={{\left( \cos A+\sin A \right)}^{2}}$

Half Angles

From (25), $\sin A=\sqrt{\frac{1-\cos 2A}{2}}$

(29) $\sin \left( \frac{A}{2} \right)=\sqrt{\frac{1-\cos A}{2}}$

(30) $\cos \left( \frac{A}{2} \right)=\sqrt{\frac{1+\cos A}{2}}$

(31) $\tan \left( \frac{A}{2} \right)=\sqrt{\frac{1-\cos A}{1+\cos A}}$