Question

Show that :
(i) $sinAsin(B—C)+sinBsin(C—A)+sinCsin(A—B)=0$
(ii) $sin(B—C)cos(A—D)+sin(C—A)cos(B—D)+sin(A—B)cos(C—D)=0$

Answer 1

(i) We have,
LHS$=sinAsin(B—C)+sinBsin(C—A)+sinCsin(A—B)=\frac{1}{2}\left[2sinAsin\right(B—C)+2sinBsin(C—A)+2sinCsin(A—B\left)\right)=\frac{1}{2}\left[cos\right(A—B+C)—cos(A+B—C)+cos(B—C+A)—cos(B+C—A)+cos(C—A+B)—cos(C+A—B\left)\right]=\frac{1}{2}\left[cos\right(A—B+C)—cos(A—B+C)—cos(A+B—C)+cos(A+B—C)—cos(B++C—A)+cos(B+C—A\left)\right]$
$=\frac{1}{2}\times 0$
= RHS
$\therefore sinAsin(B—C)+sinBsin(C—A)+sinCsin(A—B)=0$

(ii) We have,
$LHS=sin(B—C)cos(A—D)+in(C—A)cos(B—D)+sin(A—B)cos(C-D)=\frac{1}{2}\left[sin\right(B—C\left)cos\right(A—D)+2sin(C—A\left)cos\right(B—D)+2sin(A-B\left)cos\right(C—D\left)\right]=\frac{1}{2}\left[sin\right(B—C+A—D)+sin(B—C—A+D)+sin(C—A+B—D)+sin(C-A-B+D)+sin(A—B+C—D)+sin(A—B—C+D\left)\right]=\frac{1}{2}\left[sin\right(A+B—C—D)+sin(B+D—C—A)+sin(B+C—A—D)+sin(C+D-A—B)+sin(A+C—B—D)+sin(A+D—B—C\left)\right)]=\frac{1}{2}[sin(A+B—C—D)+sin(B+D—C—A)+sin\{-(A+D-B-C\left)\right\}+sin\{-(A+B-C-D\left)\right\}+sin\{-(B+D-A-C\left)\right\}+sin(A+D-B-C)]=\frac{1}{2}[sin(A+B-C-D)]+sin(B+D-C-A)-sin(A+D-B-C)-sin(A+B-C-D)-sin(B+D-A-C)+sin(A+D-B-C)=\frac{1}{2}\times 0[\because sin(-\theta )=-sin\theta ]=0=RHS\therefore sin(B-C\left)cos\right(A-D)+sin(C-A\left)cos\right(B-D)+sin(A-B\left)cos\right(C-D)=0$