# What is the period of f(theta) = tan ( ( 13 theta)/4 )- cos ( ( 2 theta)/ 5 ) ?

##### 1 Answer
Jul 7, 2016

$f$ is periodic and its Principal Period is $20 \pi .$

#### Explanation:

We know that the Principal Period of $\tan$ & $\cos$ are, $\pi$ & $2 \pi$, resp., which means that,

$\forall x \in \mathbb{R} , \tan \left(x + \pi\right) = \tan x \ldots \ldots \ldots . . \left(1\right) ,$ and, $\cos \left(x + 2 \pi\right) = \cos x \ldots \ldots \ldots \ldots . \left(2\right)$

We will use this fact in solving this problem.

Put $g \left(\theta\right) = \tan \left(13 \frac{\theta}{4}\right)$ and $h \left(\theta\right) = \cos \left(2 \frac{\theta}{5}\right)$ to give $f \left(\theta\right) = g \left(\theta\right) - h \left(\theta\right)$

By $\left(1\right) , \tan \left(13 \frac{\theta}{4}\right) = \tan \left(13 \frac{\theta}{4} + \pi\right) = \tan \left\{\frac{13}{4} \left(\theta + 4 \frac{\pi}{13}\right)\right\} ,$, i.e.,
$g \left(\theta\right) = g \left(\theta + 4 \frac{\pi}{13}\right) ,$ hence, the period ${p}_{1}$ of fun. $g$ is ${p}_{1} = 4 \frac{\pi}{13}$

On the same line, we can get the period ${p}_{2}$ of fun. $h , {p}_{2} = 2 \frac{\pi}{\frac{2}{5}} = 5 \pi .$

If we can find $l , m \in \mathbb{N}$ such that $l \cdot {p}_{1} = m \cdot {p}_{2} = p \ldots \ldots \left(3\right)$,then $f = g - h$ will be periodic and $p$ will be its Prin. Period.

Now, $l \cdot {p}_{1} = m \cdot {p}_{2} \Rightarrow l \left(4 \frac{\pi}{13}\right) = m \left(5 \pi\right) \Rightarrow 4 l = 65 m .$ Therefore, taking $l = 65 , \mathmr{and} , m = 4$, we see that $l , m \in \mathbb{N} , \mathmr{and} , \left(3\right)$ is satisfied, bcz., $65 \cdot {p}_{1} = 4 \cdot {p}_{2} , = 65 \cdot 4 \frac{\pi}{13} = 4 \cdot 5 \pi = 20 \pi = p$

Therefore, $f = g - h$ is periodic with Prin. Prd. $p = 20 \pi .$

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