by using demover theorem prove that sin2θ=2sinθcosθ


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Question Number 130103 by mohammad17 last updated on 22/Jan/21

$b y u s i n g d e m o v e r t h e o r e m p r o v e t h a t$ $s i n 2 θ = 2 s i n θ c o s θ$
	Answered by Olaf last updated on 22/Jan/21
	$(cosθ+isinθ)^n = cos(nθ)+isin(nθ) For n = 2 : (cosθ+isinθ)^2 = cos(2θ)+isin(2θ) cos^2 θ−sin^2 θ+2isinθcosθ = cos(2θ)+isin(2θ) ⇒ { ((cos^2 θ−sin^2 θ = cos(2θ))),((2sinθcosθ = sin(2θ))) :}$
	${(\cos θ + i \sin θ)}^{n} = \cos (n θ) + i \sin (n θ)$ $For n = 2 :$ ${(\cos θ + i \sin θ)}^{2} = \cos (2 θ) + i \sin (2 θ)$ $\cos^{2} θ - \sin^{2} θ + 2 i \sin θ \cos θ = \cos (2 θ) + i \sin (2 θ)$ $\Rightarrow {\begin{cases} \cos^{2} θ - \sin^{2} θ = \cos (2 θ) \\ 2 \sin θ \cos θ = \sin (2 θ) \end{cases}$
	Commented by mohammad17 last updated on 23/Jan/21

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