Question

What is the period of function ${\sin ^4}x + {\cos ^2}x$?

Answer 1

To find the period of function $f(x) = {\sin ^4}x + {\cos ^2}x$, we will be using the concept of function to solve the problem. As we know, a periodic function is a function that repeats its values after every particular interval. Using the periodicity of sine and cosine functions we will check for the smallest value for which $f(T + x) = f(x)$ by putting $T = \dfrac{\pi }{2}$.

Complete step by step answer:
We have been given a function $f(x) = {\sin ^4}x + {\cos ^2}x$ and we have to find the period of $f(x)$.
We know that periodic functions are those which repeat their values after a fixed constant interval called the period.
Generally, for a function $f(x)$, if $f(T + x) = f(x)$ then $T$ is the period.
For example, the sine function has a period of $2\pi$ because $2\pi$ is the smallest value for which the value of $\sin \left( {2\pi + x} \right) = \sin x$, for all values of $x$.
Now, we have to find the period of $f(x) = {\sin ^4}x + {\cos ^2}x$.
$\Rightarrow f(T + x) = {\sin ^4}\left( {T + x} \right) + {\cos ^2}\left( {T + x} \right)$
Putting $T = \dfrac{\pi }{2}$, we see that,
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\sin ^4}\left( {\dfrac{\pi }{2} + x} \right) + {\cos ^2}\left( {\dfrac{\pi }{2} + x} \right)$
As we know that, $\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x$ and $\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x$.
Putting this, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\left( {\cos x} \right)^4} + {\left( { - \sin x} \right)^2}$
On simplification, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^4}x + {\sin ^2}x$
On rewriting, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x \times {\cos ^2}x + {\sin ^2}x$
As we know, ${\cos ^2}x + {\sin ^2}x = 1$ i.e., ${\cos ^2}x = 1 - {\sin ^2}x$. Using this, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x\left( {1 - {{\sin }^2}x} \right) + {\sin ^2}x$
On multiplication, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x - {\cos ^2}x{\sin ^2}x + {\sin ^2}x$
On taking common, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x + {\sin ^2}x\left( {1 - {{\cos }^2}x} \right)$
Using ${\cos ^2}x + {\sin ^2}x = 1$, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x + {\sin ^2}x \times \left( {{{\sin }^2}x} \right)$
On simplifying, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\cos ^2}x + {\sin ^4}x$
On rewriting, we get
$\Rightarrow f\left( {\dfrac{\pi }{2} + x} \right) = {\sin ^4}x + {\cos ^2}x$
$= f\left( x \right)$
Therefore, the period of function ${\sin ^4}x + {\cos ^2}x$ is $\dfrac{\pi }{2}$.

Note:
Here, for $f(x) = {\sin ^4}x + {\cos ^2}x$, $f(\pi + x) = f(x)$. But, $\pi$ is not the period of $f(x)$ because the period is always the least value of $T$ which satisfies $f(T + x) = f(x)$. To solve this type of question, we need to know the fundamental period of trigonometric functions. Generally, we have three basic trigonometric functions, sine, cosine and tan having periods $2\pi$, $2\pi$ and $\pi$ respectively.