Consider the following for the next items that follow: A function is defined by f(x) = π + sin2 x.
What is the period of the function?
Finding the Period of a Trigonometric Function
The question asks for the period of the function defined by \(f(x) = \pi + \sin^2 x\). Understanding the concept of the period of a function is crucial for solving this problem.
What is the Period of a Function?
A function \(f(x)\) is said to be periodic if there exists a positive constant \(T\) such that \(f(x + T) = f(x)\) for all values of \(x\) in the domain of \(f\). The smallest such positive constant \(T\) is called the fundamental period or simply the period of the function.
Analyzing the Given Function \(f(x) = \pi + \sin^2 x\)
The function \(f(x)\) is a sum of a constant term, \(\pi\), and a trigonometric term, \(\sin^2 x\).
- The constant term \(\pi\) does not affect the period of the function. If a function \(g(x)\) has a period \(T\), then \(g(x) + c\) (where \(c\) is a constant) also has the same period \(T\).
- Therefore, the period of \(f(x) = \pi + \sin^2 x\) is determined solely by the period of the term \(\sin^2 x\).
Determining the Period of \(\sin^2 x\)
To find the period of \(\sin^2 x\), we can use trigonometric identities. A useful identity relating \(\sin^2 x\) to a function of \(2x\) is:
\(\sin^2 x = \frac{1 - \cos(2x)}{2}\)
Now substitute this into the function \(f(x)\):
\(f(x) = \pi + \frac{1 - \cos(2x)}{2}\)
\(f(x) = \pi + \frac{1}{2} - \frac{1}{2}\cos(2x)\)
\(f(x) = \left(\pi + \frac{1}{2}\right) - \frac{1}{2}\cos(2x)\)
This rewritten form shows that \(f(x)\) is a sum of a constant term \(\left(\pi + \frac{1}{2}\right)\) and a scaled cosine term \(-\frac{1}{2}\cos(2x)\). The constant term does not affect the period. The scaling factor \(-\frac{1}{2}\) also does not affect the period.
The period is determined by the argument of the cosine function, which is \(2x\).
For a function of the form \(A\cos(Bx + C) + D\), the period is given by \(\frac{2\pi}{|B|}\).
In our case, the relevant term is \(-\frac{1}{2}\cos(2x)\), which is of the form \(A\cos(Bx)\) with \(A = -\frac{1}{2}\) and \(B = 2\). The period of \(\cos(2x)\) is:
Period \(T = \frac{2\pi}{|B|} = \frac{2\pi}{|2|} = \frac{2\pi}{2} = \pi\)
Since the period of \(\sin^2 x\) is \(\pi\), and the constant term \(\pi\) does not change the period, the period of \(f(x) = \pi + \sin^2 x\) is also \(\pi\).
Conclusion
The period of the function \(f(x) = \pi + \sin^2 x\) is \(\pi\).
Let's check the options:
- \(2\pi\)
- \(\pi\)
- \(\frac{\pi}{2}\)
- The function is non-periodic
The calculated period matches the second option, \(\pi\).
| Function | Period |
|---|---|
| \(\sin(Bx)\) | \(\frac{2\pi}{|B|}\) |
| \(\cos(Bx)\) | \(\frac{2\pi}{|B|}\) |
| \(\tan(Bx)\) | \(\frac{\pi}{|B|}\) |
Revision Table: Function Period Concepts
| Concept | Description | Effect on Period |
|---|---|---|
| Periodic Function | \(f(x+T) = f(x)\) for smallest \(T > 0\) | Defines the recurring interval |
| Adding a Constant \(c\) | \(g(x) = f(x) + c\) | No change in period |
| Multiplying by a Constant \(a\) | \(g(x) = a \cdot f(x)\) | No change in period (if \(a \neq 0\)) |
| Composition \(g(x) = f(Bx)\) | Argument is multiplied by \(B\) | Period becomes \(\frac{T_{original}}{|B|}\) |
Additional Information: Periodicity and Transformations
When dealing with trigonometric functions and finding their periods, it's important to remember how transformations affect the period. A horizontal stretch or compression changes the period.
For example, the period of \(\cos(x)\) is \(2\pi\). The period of \(\cos(2x)\) is \(\frac{2\pi}{2} = \pi\). The period of \(\cos(\frac{x}{2})\) is \(\frac{2\pi}{1/2} = 4\pi\).
Using trigonometric identities to rewrite functions into simpler forms involving basic sines or cosines with linear arguments (like \(Bx\)) is a standard technique for finding periods of more complex trigonometric expressions like \(\sin^2 x\) or \(\cos^2 x\).
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