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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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