Question

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

Answer 1

Hint: 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.