以下是对第一册中推导的公式的练习，这些公式用于将两个正弦或两个余弦的和或差转换为积。
[1] sin 7 θ − sin 5 θ cos 7 θ + cos 5 θ = tan θ {\displaystyle {\frac {\sin 7\theta -\sin 5\theta }{\cos 7\theta +\cos 5\theta }}=\tan \theta }
[2] cos 6 α − cos 4 α sin 6 α + sin 4 α = − tan α {\displaystyle {\frac {\cos 6\alpha -\cos 4\alpha }{\sin 6\alpha +\sin 4\alpha }}=-\tan \alpha }
[3] sin A + sin 3 A cos A + cos 3 A = tan 2 A {\displaystyle {\frac {\sin A+\sin 3A}{\cos A+\cos 3A}}=\tan 2A}
[4] sin 7 A − sin A sin 8 A − sin 2 A = cos 4 A sec 5 A {\displaystyle {\frac {\sin 7A-\sin A}{\sin 8A-\sin 2A}}=\cos 4A\sec 5A}
[5] cos 2 ϕ + cos 2 θ cos 2 ϕ − cos 2 θ = cot ( ϕ + θ ) cot ( ϕ − θ ) {\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] sin 2 A + sin 2 B sin 2 A − sin 2 B = tan ( A − B ) cot ( A − B ) {\displaystyle {\frac {\sin 2A+\sin 2B}{\sin 2A-\sin 2B}}=\tan \left(A-B\right)\cot \left(A-B\right)}
[7] sin A + sin 2 A cos A − cos 2 A = cot ( A 2 ) {\displaystyle {\frac {\sin A+\sin 2A}{\cos A-\cos 2A}}=\cot \left({\frac {A}{2}}\right)}
[8] sin 5 λ − sin 3 λ cos 5 λ + cos 3 λ = tan λ {\displaystyle {\frac {\sin 5\lambda -\sin 3\lambda }{\cos 5\lambda +\cos 3\lambda }}=\tan \lambda }
[9] cos 2 B − cos 2 A sin 2 B + sin 2 A = tan ( A − B ) {\displaystyle {\frac {\cos 2B-\cos 2A}{\sin 2B+\sin 2A}}=\tan \left(A-B\right)}
[10] cos ( ϕ + θ ) + sin ( ϕ − θ ) = 2 sin ( 45 o + ϕ ) cos ( 45 o + θ ) {\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] sin α + sin β sin α − sin β = tan ( α + β 2 ) cot ( α − β 2 ) {\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] cos ψ + cos ω cos ω − cos ψ = cot ( ψ + ω 2 ) cot ( ψ − ω 2 ) {\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] sin ϕ + sin θ cos ϕ + cos θ = tan ( ϕ + θ 2 ) {\displaystyle {\frac {\sin \phi +\sin \theta }{\cos \phi +\cos \theta }}=\tan \left({\frac {\phi +\theta }{2}}\right)}
[14] sin A − sin B cos B − cos A = cot ( A + B 2 ) {\displaystyle {\frac {\sin A-\sin B}{\cos B-\cos A}}=\cot \left({\frac {A+B}{2}}\right)}
[15] cos 3 A − cos A sin 3 A − sin A + cos 2 A − cos 4 A sin 4 A − sin 2 A = sin A cos 2 A cos 3 A {\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] a cos ϕ + b sin ϕ = a 2 + b 2 cos [ ϕ − tan − 1 ( b a ) ] {\displaystyle a\cos \phi +b\sin \phi ={\sqrt {a^{2}+b^{2}}}\cos[\phi -\tan ^{-1}({\frac {b}{a}})]}
![{\displaystyle a\cos \phi +b\sin \phi ={\sqrt {a^{2}+b^{2}}}\cos[\phi -\tan ^{-1}({\frac {b}{a}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b82272f427130e04d9cebc778ab7ae0972c822e)
