Question

Explain how to derive a formula for sin(A - B + C)

Answer 1

$\textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta) \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(A-B+C)\implies sin[~(\stackrel{Z}{A-B})~+C]\implies sin(Z+C) \\\\\\ sin(Z)cos(C)+cos(Z)sin(C)\implies sin(A-B)cos(C)+cos(A-B)sin(C) \\\\[-0.35em] ~\dotfill\\\\ \textit{and since we know that}\\\\ sin(A-B)\implies sin(A)cos(B)-cos(A)sin(B)$

$cos(A-B)\implies cos(A)cos(B)+sin(A)sin(B)\\\\ then \\\\[-0.35em] ~\dotfill\\\\\ [sin(A)cos(B)-cos(A)sin(B)]cos(C)+[cos(A)cos(B)+sin(A)sin(B)]sin(C) \\\\\\ ~\hfill \begin{array}{cllll} sin(A)cos(B)cos(C)-cos(A)sin(B)cos(C)\\\\ +\\\\ cos(A)cos(B)sin(C)+sin(A)sin(B)sin(C) \end{array}~\hfill$