Andhra Pradesh Board Solutions for Exercise 6: Exercise 6 (f)

If $A+B+C=\frac{3\pi }{2}$, then prove that ${\cos }^{2}A+{\cos }^{2}B-{\cos }^{2}C=-2\cos A\cos B\sin C$

If $A+B+C=\frac{3\pi }{2}$, then prove that $\sin 2A+\sin 2B-\sin 2C=-4\sin A\sin B\cos C$

If $A+B+C=0$, then prove that $\sin 2A+\sin 2B+\sin 2C=-4\sin A\sin B\sin C$

If $A+B+C=0$, then prove that $\sin A+\sin B-\sin C=-4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$

If $A+B+C=2\mathrm{\pi }$, then prove that $\sin A-\sin B+\sin C-\sin D=-4\cos \frac{A+B}{2}\sin \frac{A+C}{2}\cos \frac{A+D}{2}$

If $A+B+C=2\mathrm{\pi }$, then prove that $\cos 2 \mathrm{A}+\cos 2 \mathrm{B}+\cos 2C+\cos 2\mathrm{D}=4\cos (\mathrm{A}+\mathrm{B})\cos (\mathrm{A}+\mathrm{C})\cos (\mathrm{A}+\mathrm{D})$

If $A+B+C=S$, then prove that $\sin (S-A)+\sin (S-B)+\sin C=4\cos \frac{S-A}{2}\cos \frac{S-B}{2}·\sin \frac{C}{2}$

If $A+B+C=S$, then prove that $\cos (\mathrm{S}-\mathrm{A})+\cos (\mathrm{S}-\mathrm{B})+\mathrm{cosC}=-1+4\cos \frac{\mathrm{S}-\mathrm{A}}{2}\cos \frac{\mathrm{S}-\mathrm{B}}{2}\cos \frac{\mathrm{C}}{2}$