How do I show algebraically that the period of the tangent function is $\pi$?
How do I show that the positive real number $p$ for which $\tan (x+p)=\tan (x)$ is equal to $\pi$? In essence how do I prove the period of the tangent function is $\pi$? Please bear in mind I am a pre-calculus student, thanks.
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$\begingroup$Hint: Use the identity $$\tan(\alpha + \beta)=\frac{\tan\alpha + \tan \beta}{1-\tan\alpha \tan\beta}$$
Edit:In order to prove that $\pi$ is indeed the period, we need to show that $\tan(x+\pi)=\tan(x)$. Let's start from the LHS and get to the RHS using the identity. We let $\alpha = x$ and $\beta= \pi$, also remember that $\tan \pi =0$, so: $$\tan(x + \pi)=\frac{\tan x + \tan \pi}{1-\tan x \tan\pi} = \frac{\tan x + 0}{1-(\tan x)(0)} = \frac {\tan x}{1} = \tan x$$ Thus $\pi$ is the period. By the way if you start with the problem "$\tan (x + p) = \tan x$, find the period $p$" then you would have many different solutions, because $p=2\pi$ is also a period that satisfies the equation. Infact, any $n\pi$ works as a period where $n$ is a nonzero integer.
Edit 2: Let's try to find the period $p$ instead of proving that $\pi$ is a period. The method is similar. We know that if $p$ is the period, then $\tan(x+p)=\tan x$ which is the key fact here. So we want a $p$ that satisfies this: $$\tan(x + p)=\frac{\tan x + \tan p}{1-\tan x \tan p}=\tan x$$ Manipulating this expression gives: $$\tan x + \tan p=\tan x - \tan^2 x \tan p \implies \tan p(\tan^2 x +1)=0 \implies \tan p=0 \\ \implies p=n\pi$$ for $n= ... -3,-2,-1,1,2,3 \dots$
If you put $n=1$ you get $p=\pi$ which is what we proved earlier to be a period. Also, $n \neq 0$, because $p$ can't be zero due to the definition of a period.
$\endgroup$ 5 $\begingroup$$$\tan(x+p)-\tan x=\frac{\sin(x+p)}{\cos(x+p)}-\frac{\sin x}{\cos x}=\frac{\sin(x+p-x)}{\cos(x+p)\cos x}$$
We know, if $\sin p=0\implies p=n\pi$ where $n$ is any integer
$\endgroup$ 8 $\begingroup$The function $$\tan t:={\sin t\over\cos t}$$ is on its domain the quotient of two functions with the property $f(t+\pi)\equiv-f(t)$. It follows that $\tan$ has period $\pi$.
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