# Trigonometric Functions and Equations – Transformation

## Trigonometric Functions and Equations – Transformation

i) sin (A + B) + sin (A – B) = 2 sin A cos B,

ii) sin (A + B) – sin (A – B) = 2 cos A sin B,

iii) cos (A + B) + cos (A – B) = 2 cos A cos B,

iv) cos (A – B) – cos (A + B) = 2 sin A sin B,

v) $$\sin C+\sin D=2\sin \left( \frac{C+D}{2} \right)\cos \left( \frac{C-D}{2} \right)$$,

vi) $$\sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)$$,

vii) $$\cos C+\cos D=2\cos \left( \frac{C+D}{2} \right)\cos \left( \frac{C-D}{2} \right)$$,

viii) $$\cos C-\cos D=-2\sin \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)$$.

Example: If $$\frac{\sin A-\sin C}{\cos C-\cos A}=\cot B$$, then A, B and C are in

Solution: Given that $$\frac{\sin A-\sin C}{\cos C-\cos A}=\cot B$$,

(∵ $$\cos C-\cos D=-2\sin \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)$$ & $$\sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)$$)

$$\frac{2\cos \left( \frac{A+C}{2} \right)\sin \left( \frac{A-C}{2} \right)}{2\sin \left( \frac{A+C}{2} \right)\sin \left( \frac{A-C}{2} \right)}=\cot B$$,

$$\cot \left( \frac{A+C}{2} \right)=\cot B$$,

$$\left( \frac{A+C}{2} \right)=B$$,

Hence, A and B and C will be in AP.