Question

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

Answer 1

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

Explanation:

Period of `tan (3/7theta ), P_1 = pi/(3/7) = 7/3 pi`.

Period of `- sec (5/6theta ), P_2 = (2pi)/(5/6) = 12/5`

Now,

the period of f( theta ), least possible `P = L P_1 = MP_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 `( a theta + b )`,

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

integer multiple of `( 2 pi )`.

Let `N = K L M = LCM ( L, M )`.

Multiply by the LCM of the denominators in `P_1 and P_2`

= (3)(5) = 15. Then

`15 P = L(35pi) = M(36)pi`.

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

L = 36, M = 35, and P = 84 `pi` .

Verification:

`f ( theta + 84 pi )`

`= tan ( 3/7 theta + 12 pi ) - sec ( 5/6 theta + 14 pi )`

`= tan ( 3/7 theta ) - sec ( 5/6 theta )`

`= f (theta )`

If P is halved,

`f ( theta + 42 pi ) = an ( 3/7 theta + 6 pi ) - sec ( 5/6 theta + 7 pi )`

`= tan ( 3/7 theta ) + sec ( 5/6 theta )`

`ne f( theta )`

Graph, for one period, `x in [-42pi, 42pi )` :

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