prove that sin(2x + 2h)−sin 2x = 2cos(2x +h)sin h


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Question Number 128130 by physicstutes last updated on 04/Jan/21

$prove that$ $\sin (2 x + 2 h) - \sin 2 x = 2 \cos (2 x + h) \sin h$
	Answered by Olaf last updated on 04/Jan/21

	$\sin a - \sin b = 2 \sin (\frac{a - b}{2}) \cos (\frac{a + b}{2}) (1)$ $a = 2 x + 2 h and b = 2 x$ $(1) : \sin (2 x + 2 h) - \sin (2 x) =$ $2 \sin h \cos (2 x + h)$
	Answered by Olaf last updated on 04/Jan/21

	$You can use Moivre :$ $\sin (2 x + 2 h) - \sin 2 x =$ $\frac{e^{i (2 x + 2 h)} - e^{- i (2 x + 2 h)}}{2 i} - \frac{e^{2 i x} - e^{- 2 i x}}{2 i} =$ $\frac{1}{2 i} [e^{i h} (e^{i (2 x + h)} + e^{- i (2 x + h)}) - e^{- i h} (e^{i (2 x + h)} + e^{- i (2 x + h)})] =$ $2 \frac{e^{i h} - e^{- i h}}{2 i} (\frac{e^{i (2 x + h)} + e^{- i (2 x + h)}}{2}) =$ $2 \sin h \cos (2 x + h)$
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