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Question Number 86461 by Rio Michael last updated on 28/Mar/20

Use exponential representation of \sin θ and \cos θ to show that

a) \sin^{2} θ + \cos^{2} θ = 1 b) \cos^{2} θ - \sin^{2} θ = \cos 2 θ

c) 2 \sin θ \cos θ = 2 \sin 2 θ .

Answered by TANMAY PANACEA. last updated on 28/Mar/20

c) 2 s i n θ c o s θ

= \frac{2}{i} (i s i n θ) (c o s θ)

= \frac{2}{i} \times (\frac{e^{i θ} - e^{- i θ}}{2}) (\frac{e^{i θ} + e^{- i θ}}{2})

= \frac{2}{i} \times \frac{1}{4} \times (e^{i 2 θ} - e^{- i 2 θ})

= \frac{1}{2 i} \times (2 i s i n 2 θ) = s i n 2 θ

Answered by TANMAY PANACEA. last updated on 28/Mar/20

e^{i θ} = c o s θ + i s i n θ e^{- i θ} = c o s θ - i s i n θ

c o s θ = \frac{e^{i θ} + e^{- i θ}}{2} i s i n θ = \frac{e^{i θ} - e^{- i θ}}{2}

a) {c o s}^{2} θ + {s i n}^{2} θ

= {c o s}^{2} θ - {(i s i n θ)}^{2}

= {(\frac{e^{i θ} + e^{- i θ}}{2})}^{2} - {(\frac{e^{i θ} - e^{- i θ}}{2})}^{2}

= \frac{4 \times e^{i θ} \times e^{- i θ}}{4} = 1

Commented by Rio Michael last updated on 28/Mar/20

thank you sir

Commented by peter frank last updated on 29/Mar/20

t h a n k y o u s i r

Answered by TANMAY PANACEA. last updated on 28/Mar/20

b) {c o s}^{2} θ - {s i n}^{2} θ

= {c o s}^{2} θ + {(i s i n θ)}^{2}

= {(\frac{e^{i θ} + e^{- i θ}}{2})}^{2} + {(\frac{e^{i θ} - e^{- i θ}}{2})}^{2}

= \frac{2 (e^{i 2 θ} + e^{- i 2 θ})}{4} = \frac{2 c o s 2 θ}{2} = c o s 2 θ