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 clip_image002[4]

\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)

image

=\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

clip_image002[6]

(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}}

Post Author: E-Maths