: +91 124 4007927

# TRIGONOMETRY FORMULAE

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

Posted on