Why is arctan⁡(tan⁡(25π4))=π4, and how can I get from the expression on the left to the one on the right?

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braodagxj · Accepted Answer

We can apply the identity  arctan⁡(tan⁡(x))=x  only for x∈(−π2,π2). So we just need to use the fact that  tan⁡(25π4)=tan⁡(π4)  by periodicity of the tangent.

RizerMix · Accepted Answer

The tan⁡(x) function is periodic, meaning it will not be injective, thus will not have an inverse. Unless you restrict the domain. For example, in your case tan⁡(25π4)=tan⁡(π4) So if you apply the inverse function - what value do we take? Both 25π4 and π4 give the same answer under the tangent function. So we are the ones who must decide which value we take. In exactly the same way that we decide 4=2, and not -2 (even though both (−2)2=4 as well). The decision we make is called the principle values. For example we say the principle square root of four is two. We take the principle values of arctan(x) to be from the interval (π2,π2), and since tan⁡(25π4)=tan⁡(π4) and π2≤π4≤π2, thats

alenahelenash · Accepted Answer

Since the tangent function is periodic with period π,tan(25π4)=tan(π4). And since arctan is the inverse of the restriction of tan to (−π2,π2) and since π4∈(−π2,π2), arctan⁡(tan⁡(π4))=π4

Why is \arctan(\tan(\frac{25\pi}{4}))=\frac{\pi}{4}, and how can I get from the

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