Expert Answer to "How can you show that 11−cos⁡θ−isin⁡θ=12+i2×cot⁡θ2 where i=−1?"

Answered question

2022-01-17

How can you show that

\frac{1}{1 - \cos θ - i \sin θ} = \frac{1}{2} + \frac{i}{2} \times \cot \frac{θ}{2}

where

i = \sqrt{- 1} ?

Answer & Explanation

nick1337

Expert2022-01-17Added 777 answers

Step 1
Set $ϕ = \frac{θ}{2}$
Then the expression at the left-hand side is
$\frac{1}{1 - e^{2 i ϕ}} = \frac{1}{e^{i ϕ}} \frac{1}{e^{- i ϕ} - e^{i ϕ}}$
Now we know that $e^{i ϕ} - e^{- i ϕ} = 2 i \sin ϕ$ , so we obtain
$- \frac{e^{- i ϕ}}{2 i \sin ϕ} = - \frac{\cos ϕ}{2 i \sin ϕ} + i \frac{\sin ϕ}{2 i \sin ϕ}$
which becomes
$\frac{1}{2} + \frac{i}{2} \cot ϕ$

star233

Skilled2022-01-17Added 403 answers

Step 1
$L H S$
$\Rightarrow \frac{1}{1 - \cos θ - i \sin θ} = \frac{1}{(1 - \cos θ) - i \sin θ}$
$= \frac{1}{2 \sin^{2} \frac{θ}{2} - i 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2} (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}$
$= \frac{1}{2 \sin \frac{θ}{2}} \frac{1}{(\sin \frac{θ}{2} - i \cos \frac{θ}{2})} \times \frac{(\sin \frac{θ}{2} + i \cos \frac{θ}{2})}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2})}$
$= \frac{1}{2 \sin \frac{θ}{2}} \frac{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}{1}$
$= \frac{\sin \frac{θ + i}{\cos} \frac{θ}{2}}{2 \sin \frac{θ}{2}}$
$= \frac{1}{2} \frac{\sin \frac{θ}{2}}{\sin \frac{θ}{2}} + \frac{1}{2} i \frac{\cos \frac{θ}{2}}{\sin \frac{θ}{2}}$
$= \frac{1}{2} + \frac{1}{2} i \cot \frac{θ}{2}$
$= R . H . S$

alenahelenash

Expert2022-01-24Added 556 answers

Step 1

\frac{1}{(1 - \cos θ) - i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} - 2 i \sin \frac{θ}{2} \cos \frac{θ}{2}}

= \frac{1}{2 \sin \frac{θ}{2}} \frac{1}{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}

= \frac{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}{2 \sin \frac{θ}{2}}

= \frac{1}{2} (1 + i \cot \frac{θ}{2})

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