(1)

\begin{align*} \sin A+\sin B+\sin C & =4\sin\frac{B+C}{2}\sin\frac{C+A}{2}\sin\frac{A+B}{2}+\sin\left(A+B+C\right)\\ & =4\sin\frac{\pi-A}{2}\sin\frac{\pi-B}{2}\sin\frac{\pi-C}{2}+\sin\pi\\ & =4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \end{align*}

(2)

\begin{align*} \sin A\sin B\sin C & =\frac{1}{4}\left\{ \sin\left(A+B+C\right)+\sin\left(-A+B+C\right)+\sin\left(A-B+C\right)+\sin\left(A+B-C\right)\right\} \\ & =\frac{1}{4}\left\{ \sin\pi+\sin\left(\pi-2A\right)+\sin\left(\pi-2B\right)+\sin\left(\pi-2C\right)\right\} \\ & =\frac{1}{4}\left\{ \sin\left(2A\right)+\sin\left(2B\right)+\sin\left(2C\right)\right\} \end{align*}

(2)-2

\begin{align*} \sin\left(2A\right)+\sin\left(2B\right)+\sin\left(2C\right) & =2\sin\left(A+B\right)\cos\left(A-B\right)+2\sin C\cos C\\ & =2\sin\left(\pi-C\right)\cos\left(A-B\right)+2\sin C\cos C\\ & =2\sin C\left(\cos\left(A-B\right)+\cos C\right)\\ & =2\sin C\left(2\cos\frac{A-B+C}{2}\cos\frac{A-B-C}{2}\right)\\ & =4\sin C\cos\frac{\pi-2B}{2}\cos\frac{2A-\pi}{2}\\ & =4\sin C\cos\left(\frac{\pi}{2}-B\right)\cos\left(A-\frac{\pi}{2}\right)\\ & =4\sin A\sin B\sin C \end{align*}

(3)

\begin{align*} \cos A+\cos B+\cos C & =4\cos\frac{B+C}{2}\cos\frac{C+A}{2}\cos\frac{A+B}{2}-\cos\left(A+B+C\right)\\ & =4\cos\frac{\pi-A}{2}\cos\frac{\pi-B}{2}\cos\frac{\pi-C}{2}-\cos\pi\\ & =1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \end{align*}

(4)

\begin{align*} \cos A\cos B\cos C & =\frac{1}{4}\left\{ \cos\left(A+B+C\right)+\cos\left(-A+B+C\right)+\cos\left(A-B+C\right)+\cos\left(A+B-C\right)\right\} \\ & =\frac{1}{4}\left\{ \cos\pi+\cos\left(\pi-2A\right)+\cos\left(\pi-2B\right)+\cos\left(\pi-2C\right)\right\} \\ & =-\frac{1}{4}\left\{ 1+\cos\left(2A\right)+\cos\left(2B\right)+\cos\left(2C\right)\right\} \tag{(*)}\\ & =-\frac{1}{4}\left\{ 1+2\cos^{2}A-1+2\cos^{2}B-1+2\cos^{2}C-1\right\} \\ & =\frac{1}{2}\left\{ 1-\left(\cos^{2}A+\cos^{2}B+\cos^{2}C\right)\right\} \end{align*}

(5)

\begin{align*} 0 & =\tan\pi\\ & =\tan\left(A+B+C\right)\\ & =\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A} \end{align*} これより、
\[ \tan a+\tan b+\tan c=\tan a\tan b\tan c \]

(5)-2

\begin{align*} \tan A+\tan B+\tan C & =\tan\left(A+B\right)\left(1-\tan A\tan B\right)+\tan C\\ & =\tan\left(\pi-C\right)\left(1-\tan A\tan B\right)+\tan C\\ & =-\tan\left(C\right)\left(1-\tan A\tan B\right)+\tan C\\ & =\tan A\tan B\tan C \end{align*}

(6)

\begin{align*} 0 & =\tan^{-1}\left(\frac{\pi}{2}\right)\\ & =\tan^{-1}\left(\frac{A+B+C}{2}\right)\\ & =\frac{\tan^{-1}\frac{A}{2}+\tan^{-1}\frac{B}{2}+\tan^{-1}\frac{C}{2}-\tan^{-1}\frac{A}{2}\tan^{-1}\frac{B}{2}\tan^{-1}\frac{C}{2}}{1-\tan^{-1}\frac{A}{2}\tan^{-1}\frac{B}{2}-\tan^{-1}\frac{B}{2}\tan^{-1}\frac{C}{2}-\tan^{-1}\frac{C}{2}\tan^{-1}\frac{A}{2}} \end{align*} これより、
\[ \tan^{-1}\frac{A}{2}+\tan^{-1}\frac{B}{2}+\tan^{-1}\frac{C}{2}=\tan^{-1}\frac{A}{2}\tan^{-1}\frac{B}{2}\tan^{-1}\frac{C}{2} \]