# Show that:

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

Solution:

(i) Consider LHS :

$\sin A \sin (B-C)+\sin B \sin (C-A)+\sin C \sin (A-B)$

$=\frac{1}{2}[2 \sin A \sin (B-C)]+\frac{1}{2}[2 \sin B \sin (C-A)]+\frac{1}{2}[2 \sin C \sin (A-B)]$

$=\frac{1}{2}[\cos \{A-(B-C)\}-\cos \{A+(B-C)\}]+\frac{1}{2}[\cos \{B-(C-A)\}-\cos \{B+(C-A)\}]+$$\frac{1}{2}[\cos \{C-(A-B)\}-\cos \{C+(A-B)\}] =\frac{1}{2}[\cos (A-B+C)-\cos (A+B-C)]+\frac{1}{2}[\cos (B-C+A)-\cos (B+C-A)]+$$\frac{1}{2}[\cos (C-A+B)-\cos (C+A-B)]$

\begin{aligned}=& \frac{1}{2} \cos (A-B+C)-\frac{1}{2} \cos (A+B-C)+\frac{1}{2} \cos (B-C+A)-\frac{1}{2} \cos (B+C-A)+\frac{1}{2} \cos \\ &(C-A+B)-\frac{1}{2} \cos (C+A-B) \end{aligned}

$=\frac{1}{2} \cos (A-B+C)-\frac{1}{2} \cos (A+B-C)+\frac{1}{2} \cos (A+B-C)-\frac{1}{2} \cos (B+C-A)+\frac{1}{2} \cos (B+C-A)$$-\frac{1}{2} \cos (A-B+C) =0 = RHS (i) Consider LHS : \sin (B-C) \cos (A-D)+\sin (C-A) \cos (B-D)+\sin (A-B) \cos (C-D) =\frac{1}{2}[2 \sin (B-C) \cos (A-D)]+\frac{1}{2}[2 \sin (C-A) \cos (B-D)]+\frac{1}{2}[2 \sin (A-B) \cos (C-D)] =\frac{1}{2}[\sin \{(B-C)+(A-D)\}+\sin \{(B-C)-(A-D)\}]+$$\frac{1}{2}[\sin \{(C-A)+(B-D)\}+\sin \{(C-A)-(B-D)\}]+$$\frac{1}{2}[\sin \{(A-B)+(C-D)\}+\sin \{(A-B)-(C-D)\}] \frac{1}{2}[\sin (C-A+B-D)+\sin (C-A-B+D)]+\frac{1}{2}[\sin (A-B+C-D)+\sin (A-B-C+D)] =\frac{1}{2}[\sin (B-C+A-D)+\sin (B-C-A+D)]+$$\frac{1}{2}[\sin \{-(-C+A-B+D)\}+\sin \{-(-C+A+B-D)\}]+$$\frac{1}{2}[\sin \{-(-A+B-C+D)\}+\sin (A-B-C+D)]$

$=\frac{1}{2} \sin (B-C+A-D)+\frac{1}{2} \sin (B-C-A+D)-\frac{1}{2} \sin (-C+A-B+D)-\frac{1}{2} \sin (-C+A+B-D)$

$-\frac{1}{2} \sin (-A+B-C+D)+\frac{1}{2} \sin (A-B-C+D)$

= 0

= RHS

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