Key to "How to get cot⁡(θ2) from sin⁡θ1−cos⁡θ ?"

Tayyib Cornish

Answered question

2022-02-26

How to get

\cot (\frac{θ}{2})

from

\frac{\sin θ}{1 - \cos θ}

Answer & Explanation

Rebekah Irvine

Beginner2022-02-27Added 6 answers

All you need are the following formulas:

\sin (2 A) = 2 \sin (A) \cos (A)

and

\cos (2 A) = 1 - 2 \sin^{2} (A)

Now take

A = \frac{θ}{2}

to get that

\sin (θ) = 2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2})

and

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

Now plugging this into your equation, we get that

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

Mikayla Swan

Beginner2022-02-28Added 9 answers

Alternately, use these formulas:
$\cos (θ) = \frac{e^{i θ} + e^{- i θ}}{2}$ and $\sin (θ) = \frac{e^{i θ} - e^{- i θ}}{2 i}$
So
$\frac{\sin (θ)}{1 - \cos (θ)} = \frac{e^{i θ} - e^{- i θ}) / (2 i)}{(2 - e^{i θ} - e^{- i θ}) / 2}$
$= \frac{1}{i} \frac{(e^{i θ / 2} + e^{- i θ / 2}) (e^{i θ / 2} - e^{- i θ / 2})}{- (e^{i θ / 2} - e^{- θ / 2})^{2}}$
$= i \frac{2 \cos (\frac{θ}{2})}{2 i \sin (\frac{θ}{2})}$
$= \cot (\frac{θ}{2})$

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