Question

Prove that:sin(B-C) cos(A-D) +sin(C-A) cos(B-D)+sin(A-B) cos(C-D)=0

Answer 1

Use the product formula,

$\sin \alpha \cos \beta =\frac{1}{2} [ \sin (\alpha +\beta )+\sin (\alpha -\beta )]$

Using the above formula,

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

Now Adding the above 3 equations,

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

The proof is complete.