# Thread: trig question (correct me) [part 1]

1. ## trig question (correct me) [part 1]

hello
here's my function :
let $\mathbb{E} = [-\pi ,+\pi]$
$x \in \mathbb{E}$ $;$ $f(x) = \sqrt{\frac{1+cos(x)}{1-cos(x)}}$
and here's my questions :
Prove that $(\forall x \in \mathbb{D})$ $f(x) = \left |\tan (\frac{\pi}{2}-\frac{x}{2}) \right |$
Study the sign of $\tan (\frac{\pi}{2}-\frac{x}{2})$ in $]0,+\pi]$
i proceeded like this :
for the first one,we have $\forall x\in \mathbb{R},1+cos(x) = 2cos^{2}(\frac{x}{2})$ and $1-cos(x) = 2sin^{2}(\frac{x}{2})$
$\Rightarrow \sqrt{\frac{1+cos(x)}{1-cos(x)}} =$ $\sqrt{\frac{2cos^{2} (\frac{x}{2})}{2sin^{2}(\frac{x}{2})}}=$
$| \frac{cos(\frac{x}{2})}{sin(\frac{x}{2})}|$ $= \left |\frac{1}{\tan (\frac{x}{2})} \right |$ $= \left |\tan (\frac{\pi}{2}-\frac{x}{2}) \right |$
for the second one,am i correct writing ; $\forall x \in \mathbb{D},f(x) = tan(\frac{\pi}{2}-\frac{x}{2})$ ( i already studied $f$ )
thanks for everyone
by the way i'm trying to improve myself for writing proofs so if there is any small mistake please tell me

2. ## Well....

Originally Posted by Raoh

Prove that $(\forall x \in \mathbb{D})$ $f(x) = \left |\tan (\frac{\pi}{2}-\frac{x}{2}) \right |$
Study the sign of $\tan (\frac{\pi}{2}-\frac{x}{2})$ in $]0,+\pi]$
well to me it seems that interval of D is not given and we are separetely suppose to check its sign in interval ]0,+pi]
(you could have wrote - where D is given by ]0,+pi] )
else seems correct to me

3. sorry about later,i forgot to mention the interval of $\mathbb{D}$.
$\mathbb{D} =$ $[-\pi,0)\cup(0,+\pi]$