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3.Trigonometrical Ratios, Functions and Identities
normal
જો $cosA + cosB = cosC,\ sinA + sinB = sinC$ હોય તો સમીકરણ $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ =
A
$0$
B
$1$
C
$2$
D
$3$
Solution
${{\rm{e}}^{{\rm{iA}}}} + {{\rm{e}}^{{\rm{iB}}}} = (\cos A + {\rm{i}}\sin {\rm{A}}) + (\cos {\rm{B}} + {\rm{i}}\sin {\rm{B}})$
$ = \cos C + i\sin C = {e^{iC}}$
Also ${e^{ – iA}} + {e^{ – iB}} = {e^{ – iC}}$
$ \Rightarrow \frac{{{e^{iC}}}}{{{e^{i\left( {A + B} \right)}}}} = {e^{ – iC}} \Rightarrow {e^{2iC}} = {e^{i(A + B)}}$
$\Rightarrow \cos 2 C+i \sin 2 C$
$=\cos (A+B)+i \sin (A+B)$
$\therefore \sin (A+B)=\sin 2 C$
Standard 11
Mathematics
