Question

Show that:
(i) sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Answer 1

(i)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin A\sin \left(B-C\right)+\sin B\sin \left(C-A\right)+\sin C\sin \left(A-B\right) \\ =\frac{1}{2}\left[2\sin A\sin \left(B-C\right)\right]+\frac{1}{2}\left[2\sin B\sin \left(C-A\right)\right]+\frac{1}{2}\left[2\sin C\sin \left(A-B\right)\right] \\ =\frac{1}{2}\left[\cos \left\{A-\left(B-C\right)\right\}-\cos \left\{A+\left(B-C\right)\right\}\right]+\frac{1}{2}\left[\cos \left\{B-\left(C-A\right)\right\}-\cos \left\{B+\left(C-A\right)\right\}\right]+\frac{1}{2}\left[\cos \left\{C-\left(A-B\right)\right\}-\cos \left\{C+\left(A-B\right)\right\}\right] \\ =\frac{1}{2}\left[\cos \left(A-B+C\right)-\cos \left(A+B-C\right)\right]+\frac{1}{2}\left[\cos \left(B-C+A\right)-\cos \left(B+C-A\right)\right]+\frac{1}{2}\left[\cos \left(C-A+B\right)-\cos \left(C+A-B\right)\right] \\ =\frac{1}{2}\cos \left(A-B+C\right)-\frac{1}{2}\cos \left(A+B-C\right)+\frac{1}{2}\cos \left(B-C+A\right)-\frac{1}{2}\cos \left(B+C-A\right)+\frac{1}{2}\cos \left(C-A+B\right)-\frac{1}{2}\cos \left(C+A-B\right) \\ =\frac{1}{2}\cos \left(A-B+C\right)-\frac{1}{2}\cos \left(A+B-C\right)+\frac{1}{2}\cos \left(A+B-C\right)-\frac{1}{2}\cos \left(B+C-A\right)+\frac{1}{2}\cos \left(B+C-A\right)-\frac{1}{2}\cos \left(A-B+C\right) \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Consider}\mathrm{LHS}: \\ \sin \left(B-C\right)\cos \left(A-D\right)+\sin \left(C-A\right)\cos \left(B-D\right)+\sin \left(A-B\right)\cos \left(C-D\right) \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\left[2\sin \left(B-C\right)\cos \left(A-D\right)\right]+\frac{1}{2}\left[2\sin \left(C-A\right)\cos \left(B-D\right)\right]+\frac{1}{2}\left[2\sin \left(A-B\right)\cos \left(C-D\right)\right] \\ =\frac{1}{2}\left[\sin \left\{\left(B-C\right)+\left(A-D\right)\right\}+\sin \left\{\left(B-C\right)-\left(A-D\right)\right\}\right]+\frac{1}{2}\left[\sin \left\{\left(C-A\right)+\left(B-D\right)\right\}+\sin \left\{\left(C-A\right)-\left(B-D\right)\right\}\right]+\frac{1}{2}\left[\sin \left\{\left(A-B\right)+\left(C-D\right)\right\}+\sin \left\{\left(A-B\right)-\left(C-D\right)\right\}\right] \\ =\frac{1}{2}\left[\sin \left(B-C+A-D\right)+\sin \left(B-C-A+D\right)\right]+\frac{1}{2}\left[\sin \left(C-A+B-D\right)+\sin \left(C-A-B+D\right)\right]+\frac{1}{2}\left[\sin \left(A-B+C-D\right)+\sin \left(A-B-C+D\right)\right] \\ =\frac{1}{2}\left[\sin \left(B-C+A-D\right)+\sin \left(B-C-A+D\right)\right]+\frac{1}{2}\left[\sin \left\{-\left(-C+A-B+D\right)\right\}+\sin \left\{-\left(-C+A+B-D\right)\right\}\right]+\frac{1}{2}\left[\sin \left\{-\left(-A+B-C+D\right)\right\}+\sin \left(A-B-C+D\right)\right] \\ =\frac{1}{2}\sin \left(B-C+A-D\right)+\frac{1}{2}\sin \left(B-C-A+D\right)-\frac{1}{2}\sin \left(-C+A-B+D\right)-\frac{1}{2}\sin \left(-C+A+B-D\right)-\frac{1}{2}\sin \left(-A+B-C+D\right)+\frac{1}{2}\sin \left(A-B-C+D\right) \\ =0 \\ =\mathrm{RHS} \end{gathered}$$