The period of trigonometric functions

1. Apr 7, 2009

ShayanJ

Hi everyone

Could you give me a way to calculate the period of every trigonometric functions?
thanks

2. Apr 7, 2009

qntty

Let $$\text{trig}\, x$$ be any trig function and T be it's period. The period of

$$a\text{trig}\, b(x+c) + d$$

is T/b.

The period of tan and cot is $$\pi$$ and the period of the other functions (cos, sin, sec, csc) is $$2\pi$$. This can be remembered by the geometric definitions (sin is opp/hyp, cos is adj/hyp etc) by noticing that the ratio opp/hyp etc doesn't repeat until a complete revolution, but the ratio opp/adj and adj/opp does because (-opp)/(-adj) = opp/adj.

3. Apr 7, 2009

AUMathTutor

Just to be clear, I think qntty was saying that if f(x) is a function with period T, then the period of f(ax+b) equals T/a. Is this correct?

4. Apr 8, 2009

ShayanJ

and what about functions that are the sum of two or more trig functions and the ones that have trig functions as their nominator and/or denominator?

5. Apr 8, 2009

AUMathTutor

Generally, if you have two trig functions added together, the function is no longer periodic. Take for instance:

f(x) = sin(sqrt(2)PI*x) + cos(PI*x)
then
f'(x) = sqrt(2)PI*cos(sqrt(2)PI*x) - sin(PI*x)

Periodicity implies that f(x) = f(x+T) and f'(x) = f'(x+T). However, think about it... since the two periods are incommensurate, there is no T which you can multiply by two different integers to give you multiples of the periods of each individual sin/cos. To do this would be to solve the equation

t1 = 2PI/sqrt(2)PI = 2/sqrt(2) = sqrt(2)
t2 = 2PI/PI=2

T = n*t1 = m*t2

Such as to find the smallest possible pair of numbers (n, m). But since t1 and t2 are incommensurate, and since n, m are integers, this equation has no solutions.

In fact, functions such as sin(ax) + cos(bx) will have solutions iff the periods are commensurate, that is, they satisfy the equation I gave, and then to find the period, you find a T using the same equation.