by-using-demover-theorem-prove-that-sin2-2sin-cos-

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

{(\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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