simplify-A-arctan-sinx-1-cosx-

Question Number 30528 by abdo imad last updated on 22/Feb/18

s i m p l i f y A = a r c t a n (\frac{s i n x}{1 - c o s x}) .

Commented by abdo imad last updated on 23/Feb/18

w e h a v e A = a r c t a n (\frac{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}{2 {s i n}^{2} (\frac{x}{2})})

A = a r c t a n (c o t a n (\frac{x}{2})) = a r c t a n (\frac{1}{t a n (\frac{x}{2})})

c a s e 1 i f 0 < x < π t a n (\frac{x}{2}) > 0 \Rightarrow A = \frac{π}{2} - a r c t a n (t a n (\frac{x}{2}))

\Rightarrow A = \frac{π}{2} - \frac{x}{2} = \frac{π - x}{2}

c a s e 2 i f - π < x < 0 t a n (\frac{x}{2}) < 0 \Rightarrow A = - \frac{π}{2} - a r c t a n (t a n (\frac{x}{2}))

\Rightarrow A = - \frac{π}{2} - \frac{x}{2} .

Answered by prakash jain last updated on 22/Feb/18

\frac{\sin x}{1 - \cos x} = \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{{2 \sin}^{2} \frac{x}{2}} = \cot \frac{x}{2}

a r c t a n (\frac{s i n x}{1 - c o s x}) = \tan^{- 1} \cot \frac{x}{2}

Commented by prakash jain last updated on 23/Feb/18

what if x = 8 π, 9 π etc ?

\tan^{- 1} x should be from - \frac{π}{2} to \frac{π}{2}

Commented by mrW2 last updated on 23/Feb/18

\tan^{- 1} \cot \frac{x}{2} = \frac{π}{2} - \frac{x}{2} ? ? ?

Commented by mrW2 last updated on 23/Feb/18

f o r 0 < x < 2 π :

A = \frac{1}{2} (π - x)

g e n e r a l l y :

A = \frac{1}{2} (π + 2 π [\frac{x}{2 π}] - x)

e . g . :

x = 6.5 π

[\frac{x}{2 π}] = [\frac{6.5 π}{2 π}] = 3

A = \frac{1}{2} (π + 6 π - 6.5 π) = \frac{π}{4}

Facebook Tweet Pin