If cosA + cosB = cosC, sinA + sinB = sinC&nbs | vedclass

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Question

${{\rm{e}}^{{\rm{iA}}}} + {{\rm{e}}^{{\rm{iB}}}} = (\cos A + {\rm{i}}\sin {\rm{A}}) + (\cos {\rm{B}} + {\rm{i}}\sin {\rm{B}})$

$ = \cos C + i\sin C

Vedclass · Accepted Answer

$1$

Vedclass · Answer

${{m{e}}^{{m{iA}}}} + {{m{e}}^{{m{iB}}}} = (\cos A + {m{i}}\sin {m{A}}) + (\cos {m{B}} + {m{i}}\sin {m{B}})$

$ = \cos C + i\sin C = {e^{iC}}$

Also ${e^{ - iA}} + {e^{ - iB}} = {e^{ - iC}}$

$ \Rightarrow \frac{{{e^{iC}}}}{{{e^{i\left( {A + B} ight)}}}} = {e^{ - iC}} \Rightarrow {e^{2iC}} = {e^{i(A + B)}}$

$\Rightarrow \cos 2 C+i \sin 2 C$

$=\cos (A+B)+i \sin (A+B)$

$	herefore \sin (A+B)=\sin 2 C$

If $cosA + cosB = cosC,\ sinA + sinB = sinC$ then the value of expression $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ is

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