1. ## Equation

Hi,

I must show that $\displaystyle (\forall k \in \mathbb{Z}) \forall x \in ]-\frac{\pi}{2}+k\pi;\frac{\pi}{2}+k\pi[: Arctan(tanx)=x-k\pi.$

I know just that $\displaystyle Arctan(tanx)=x$ when $\displaystyle x \in ]-\frac{\pi}{2}+k\pi;\frac{\pi}{2}+k\pi[$

Can you help me???

2. Originally Posted by lehder
Hi,

I must show that $\displaystyle (\forall k \in \mathbb{Z}) \forall x \in ]-\frac{\pi}{2}+k\pi;\frac{\pi}{2}+k\pi[: Arctan(tanx)=x-k\pi.$

I know just that $\displaystyle Arctan(tanx)=x$ when $\displaystyle x \in ]-\frac{\pi}{2}{\color{red}+k\pi};\frac{\pi}{2}{\col or{red}+k\pi}[$
No, I think that's wrong (and contradicts what you are required to prove). Instead, you know that $\displaystyle Arctan(tanx)=x$ when $\displaystyle x \in ]-\frac{\pi}{2};\frac{\pi}{2}[$

And then you know something more: you know that $\displaystyle \tan$ has period $\displaystyle \pi$, therefore you can add / subtract $\displaystyle \pi$ an integral number of times until the value of an $\displaystyle x\in]-\frac{\pi}{2}+k\pi;\frac{\pi}{2}+k\pi[$ actually lies in $\displaystyle ]-\frac{\pi}{2};+\frac{\pi}{2}[$.