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Question Number 68964 by ahmadshah last updated on 17/Sep/19

Commented by mind is power last updated on 17/Sep/19

$m + i n = e^{i a} + e^{i b} \Rightarrow m^{2} - n^{2} + 2 i m n = e^{i 2 a} + e^{2 i b} + 2 e^{i (a + b)}$ $\Rightarrow I m (m^{2} - n^{2} + 2 i m n) = I m (e^{2 i a} + e^{2 i b} + 2 e^{i (a + b)})$ $\Rightarrow 2 m n = s i n (2 a) + s i n (2 b) + 2 s i n (a + b)$ $\Rightarrow 2 m n = 2 c o s (a - b) s i n (a + b) + 2 s i n (a + b)$ $\Rightarrow m n = s i n (a + b) (c o s (a - b) + 1)$ $∣ m + i n ∣^{2} =∣ e^{i a} + e^{i b} ∣^{2} \Rightarrow m^{2} + n^{2} = {(c o s (a) + c o s (b))}^{2} + {(s i n (a) + s i n (b))}^{2}$ $\Rightarrow m^{2} + n^{2} = 2 + 2 c o s (a) c o s (b) + 2 s i n (a) s i n (b) \Rightarrow m^{2} + n^{2} = 2 (1 + c o s (a - b))$ $\Rightarrow 1 + c o s (a - b) = \frac{m^{2} + n^{2}}{2}$ $\Rightarrow m n = s i n (a + b) (\frac{m^{2} + n^{2}}{2})$ $\Rightarrow s o n (a + b) = \frac{2 m n}{m^{2} + n^{2}}$
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