# 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.