In Calculus, we use the notion of periodic Functions frequently. A periodic function can be defined as a function whose values reiterate in recurrent spaces or intervals. There are a large number of examples of the periodic Functions that include the trigonometric functions (i.e. sine, cosine, tangent, and cotangent and secant functions). A notable feature of the functions is that they reiterate over periods of 2π radians. Periodic functions are applied all over the science and Math to depict phenomenon that display periodicity such as waves.
To understand the concept of periodic functions, let us take a function f (a). If the function is a periodic function, then there will exist some value h such that f (a + h) = f (a). For example, let us see the sine function.
Sin (a + 2 π) = sin a.
This means that sin (0 + 2 π) = sin (0) = 0, or sin (π / 3 + 2 π) = sin (π / 3) = √3 / 2. To illustrate it further we can say, if the value of a function at Point a is some ‘b’ then for it to exhibit periodicity, there will be a quantity h such that the value of the function at point a + h will also be b. The value h is termed as Period of the function. Thus, the period of a function is a value after which it repeats its behavior.
Based on the number of periods, periodic functions can be classified into singly periodic functions, doubly periodic function, and triply periodic function and so on. The fundamental or prime period of a function is the extent of a least uninterrupted part of the Domain for one complete cycle of the function. In other words, it is the smallest period that can be possible for the given function. For the sine function, the fundamental period is 2π.
Values of Trigonometric Functions can be evaluated from right angled triangle. The secant of a function is represented as follows:
Sec x = Hypotenuse (Longest side) / Base of the triangle (shorter side).... equation 1
We can use above formula to calculate other trigonometric functions. That is, value of sec x is given; rest of the functions values can be calculated. Let us suppose an example of this to understand solving Trigonometric functions.
Example: If sec x equal p plus 1 by 4p, then how to find the value of sec x plus tan x.
Solution: In the given problem we have the value of sec x = (p + 1) / 4p and we have to calculate the value of sec x + tan x.
For this we first have to calculate the value of tan x using the value of sec x. On comparing the value of sec x i.e. (p + 1) / 4p we see that hypotenuse or the longest side of the Right Triangle is equals to (p + 1) and base or one of the two shorter sides is equals to 4p. We can also use direct trigonometric identity:
Sec2 x = 1 + tan2 x,
((p + 1) / 4p) 2 = 1 + tan2 x,
(p2 + 1 + 2p) /16p2 – 1 =tan2 x,
Or tan x = √ (-15p2 + 2p + 1) / 4p,
Or we can also use pythagoras theorem to find the value of tan x.
Perpendicular side2 = Hypotenuse2 – Base2 = p2 + 1 + 2p - 16p2 = -15p2 + 2p + 1,
Or Perpendicular side = √ (-15p2 + 2p + 1),
Value of tan x = √ (-15p2 + 2p + 1) / 4p.