Answer:
Option C - [tex]F(x)=\cos(x+\pi)[/tex]
Step-by-step explanation:
To find : Which of the following options is an even function?
Solution :
To determine the function is even,odd or neither we will find f(-x)
If f(-x) = - f(x) then function is odd.
If f(-x) = f(x) then function is even.
We know that,
[tex]\sin(-x)=-\sin x[/tex] is always a odd function.
[tex]\cos(-x)=\cos x[/tex] is always an even function.
Now, In the following options
A. [tex]F(x)=\sin(x+\pi )[/tex]
[tex]F(-x)=\sin(-(x+\pi))=-\sin (x+\pi)=-F(x)[/tex]
It is a odd function.
B. [tex]F(x)=2sin(2x)[/tex]
[tex]F(-x)=2\sin(-2x)=-2\sin (2x)=-F(x)[/tex]
It is a odd function.
C. [tex]F(x)=\cos(x+\pi)[/tex]
[tex]F(-x)=\cos(-(x+\pi))=\cos(x+\pi)=F(x)[/tex]
It is an even function.
D. [tex]F(x)=2\sin(\pi+\pi)[/tex]
[tex]F(-x)=2\sin(-2x)=-2\sin (2x)=-F(x)[/tex]
It is a odd function.
Therefore, Option C is correct.
