These are exercises on the formulae derived in Book 1 for converting the sum or difference of two sines or two cosines into a product.

[1] ${\displaystyle {\frac {\sin 7\theta -\sin 5\theta }{\cos 7\theta +\cos 5\theta }}=\tan \theta }$

[2] ${\displaystyle {\frac {\cos 6\alpha -\cos 4\alpha }{\sin 6\alpha +\sin 4\alpha }}=-\tan \alpha }$

[3] ${\displaystyle {\frac {\sin A+\sin 3A}{\cos A+\cos 3A}}=\tan 2A}$

[4] ${\displaystyle {\frac {\sin 7A-\sin A}{\sin 8A-\sin 2A}}=\cos 4A\sec 5A}$

[5] ${\displaystyle {\frac {\cos 2\phi +\cos 2\theta }{\cos 2\phi -\cos 2\theta }}=\cot \left(\phi +\theta \right)\cot \left(\phi -\theta \right)}$

[6] ${\displaystyle {\frac {\sin 2A+\sin 2B}{\sin 2A-\sin 2B}}=\tan \left(A-B\right)\cot \left(A-B\right)}$

[7] ${\displaystyle {\frac {\sin A+\sin 2A}{\cos A-\cos 2A}}=\cot \left({\frac {A}{2}}\right)}$

[8] ${\displaystyle {\frac {\sin 5\lambda -\sin 3\lambda }{\cos 5\lambda +\cos 3\lambda }}=\tan \lambda }$

[9] ${\displaystyle {\frac {\cos 2B-\cos 2A}{\sin 2B+\sin 2A}}=\tan \left(A-B\right)}$

[10] ${\displaystyle \cos \left(\phi +\theta \right)+\sin \left(\phi -\theta \right)=2\sin \left(45^{o}+\phi \right)\cos \left(45^{o}+\theta \right)}$

[11] ${\displaystyle {\frac {\sin \alpha +\sin \beta }{\sin \alpha -\sin \beta }}=\tan \left({\frac {\alpha +\beta }{2}}\right)\cot \left({\frac {\alpha -\beta }{2}}\right)}$

[12] ${\displaystyle {\frac {\cos \psi +\cos \omega }{\cos \omega -\cos \psi }}=\cot \left({\frac {\psi +\omega }{2}}\right)\cot \left({\frac {\psi -\omega }{2}}\right)}$

[13] ${\displaystyle {\frac {\sin \phi +\sin \theta }{\cos \phi +\cos \theta }}=\tan \left({\frac {\phi +\theta }{2}}\right)}$

[14] ${\displaystyle {\frac {\sin A-\sin B}{\cos B-\cos A}}=\cot \left({\frac {A+B}{2}}\right)}$

[15] ${\displaystyle {\frac {\cos 3A-\cos A}{\sin 3A-\sin A}}+{\frac {\cos 2A-\cos 4A}{\sin 4A-\sin 2A}}={\frac {\sin A}{\cos 2A\cos 3A}}}$

[16] ${\displaystyle a\cos \phi +b\sin \phi ={\sqrt {a^{2}+b^{2}}}\cos[\phi -\tan ^{-1}({\frac {b}{a}})]}$