lineariser sin^5 (x)


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Question Number 176379 by doline last updated on 17/Sep/22

$l i n e a r i s e r {s i n}^{5} (x)$
	Answered by a.lgnaoui last updated on 17/Sep/22

	$\sin^{5} (x) = {(\frac{e^{i x} - e^{- i x}}{2})}^{5}$ $= \frac{1}{2^{5}} ([e^{5 i x} - 5 e^{4 i x} e^{- i x} + 10 e^{3 i x} e^{- 2 i x} - 10 e^{2 i x} e^{- 3 i x} + 5 e^{i x} e^{- 4 i x} - e^{- 5 i x})$ $= \frac{1}{2^{5}} [(e^{5 i x} - 5 e^{3 i x} + 10 e^{i x} - 10 e^{- i x} + 5 e^{- 3 i x} - e^{- 5 i x})$ $= \frac{1}{2^{5}} [(e^{5 i x} - e^{- 5 i x}) - 5 (e^{3 i x} - e^{- 3 i x}) + 10 (e^{i x} - e^{- i x})]$ $= \frac{1}{2^{4}} (\sin^{5} (x) - 5 \sin^{3} (x) + 10 \sin (x)]$
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