Respuesta :
To answer this question, we must know some identities:
1. cos(x) is an even function, so cos(x)=cos(-x) [this makes choice (c) true]
2. sin(x) and cos(x) are the same periodic functions with a phase-shift of pi/2, so that sin(x+pi/2)=cos(x) [this makes choice (b) true]
3. also, sin(x) is symmetrical about pi/2, and cos(x) is symmetrical about x=0. This means that sin(x)=cos(pi/2-x) [ this case is not present in the choices ]
1. cos(x) is an even function, so cos(x)=cos(-x) [this makes choice (c) true]
2. sin(x) and cos(x) are the same periodic functions with a phase-shift of pi/2, so that sin(x+pi/2)=cos(x) [this makes choice (b) true]
3. also, sin(x) is symmetrical about pi/2, and cos(x) is symmetrical about x=0. This means that sin(x)=cos(pi/2-x) [ this case is not present in the choices ]
Answer:
Option b and c is correct.
Step-by-step explanation:
Given : Function [tex]y=4\cos x-2[/tex]
To find : Which of the following are equivalent to the function?
Solution :
The function given is [tex]y=4\cos x-2[/tex]
We know that,
[tex]\cos(-x)=\cos(x)[/tex]
and [tex]\sin(\frac{\pi}{2}+x)=\cos(x)[/tex]
Applying these in the given function,
Using, [tex]\cos(-x)=\cos(x)[/tex]
[tex]y=4\cos (-x)-2[/tex]
[tex]y=4\cos (x)-2[/tex]
So option c is equivalent.
Using,[tex]\sin(\frac{\pi}{2}+x)=\cos(x)[/tex]
Substitute
[tex]y=4\sin(\frac{\pi}{2}+x)-2[/tex]
[tex]y=4\cos (x)-2[/tex]
So option b is equivalent.
Therefore, Option b and c is correct.