# What is the period of f(theta) = tan ( ( 3 theta)/7 )- sec ( ( 5 theta)/ 6 ) ?

Jul 29, 2018

$84 \pi$.
If necessary, I would again edit my answer myself, for debugging.

#### Explanation:

Period of $\tan \left(\frac{3}{7} \theta\right) , {P}_{1} = \frac{\pi}{\frac{3}{7}} = \frac{7}{3} \pi$.

Period of $- \sec \left(\frac{5}{6} \theta\right) , {P}_{2} = \frac{2 \pi}{\frac{5}{6}} = \frac{12}{5}$

Now,

the period of f( theta ), least possible $P = L {P}_{1} = M {P}_{2}$. So,

P = (7/3pi)L = (12/5pi)M.

If there is at least one term in the form

sine, cosine, csc or sec of $\left(a \theta + b\right)$,

P = least possible ( P/2 not the period ).

integer multiple of $\left(2 \pi\right)$.

Let $N = K L M = L C M \left(L , M\right)$.

Multiply by the LCM of the denominators in ${P}_{1} \mathmr{and} {P}_{2}$

= (3)(5) = 15. Then

$15 P = L \left(35 \pi\right) = M \left(36\right) \pi$.

As 35 and 36 are co-prime K = 1, N = (35)(36),

L = 36, M = 35, and P = 84 $\pi$ .

Verification:

$f \left(\theta + 84 \pi\right)$

$= \tan \left(\frac{3}{7} \theta + 12 \pi\right) - \sec \left(\frac{5}{6} \theta + 14 \pi\right)$

$= \tan \left(\frac{3}{7} \theta\right) - \sec \left(\frac{5}{6} \theta\right)$

$= f \left(\theta\right)$

If P is halved,

$f \left(\theta + 42 \pi\right) = a n \left(\frac{3}{7} \theta + 6 \pi\right) - \sec \left(\frac{5}{6} \theta + 7 \pi\right)$

$= \tan \left(\frac{3}{7} \theta\right) + \sec \left(\frac{5}{6} \theta\right)$

$\ne f \left(\theta\right)$

Graph, for one period, $x \in \left[- 42 \pi , 42 \pi\right)$ :

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