Prove that $sin (A - B) = sin A cos B - cos A sin B$

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Solution

We can prove this by using euler’s identity : $e^{i x} = c o s x + i s i n x$
therefore we have $e^{i A} = c o s A + i s i n A$ ,
and $e^{i B} = c o s B + i s i n B$
now
$e^{i A} * e^{i B} = e^{i (A + B)}$
$\Rightarrow (c o s A + i s i n A) (c o s B + i s i n B) = (c o s (A + B) + i s i n (A + B))$
$\Rightarrow (c o s A * c o s B - s i n A * s i n B) + i (s i n A * c o s B + c o s A * s i n B) = (c o s (A + B) + i s i n (A + B))$
On equating the imaginary parts on both sides of the equation we get the required result
$i . e . s i n (A + B) = s i n A * c o s B + c o s A * s i n B$

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