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.