How do you know if #v(x) = 2 sin x cos x# is an even or odd function?

Question

Caleb Alexander · Accepted Answer

#v(x)# it's an odd function

A function #f(x)# it's EVEN when: #f(-x)=f(x)# A function #f(x)# it's ODD when: #f(-x)=-f(x) # Now evaluate #v(-x)# #v(-x)=2sin(-x)*cos(-x)=# #=2*[-sin(x)]*cos(x)=-2sinxcosx# #:.v(-x)=-v(x) => v(x)# Odd

Sadie Alexander · Answer

undefined To determine if the function $ v(x) = 2 \sin(x) \cos(x) $ is even or odd, we need to check whether it satisfies the properties of even or odd functions.

1. Even functions satisfy the property $ f(-x) = f(x) $ for all $ x $ in their domain.
2. Odd functions satisfy the property $ f(-x) = -f(x) $ for all $ x $ in their domain.

For the function $ v(x) = 2 \sin(x) \cos(x) $, let's check these properties:

1. Substitute $ -x $ into the function:

$$ v(-x) = 2 \sin(-x) \cos(-x) $$

Since sine and cosine are both even functions, $ \sin(-x) = -\sin(x) $ and $ \cos(-x) = \cos(x) $.

So,

$$ v(-x) = 2 (-\sin(x)) \cos(x) = -2 \sin(x) \cos(x) $$

This does not equal $ v(x) $, so the function is not even.

2. Now, let's check if it's odd:

$$ v(-x) = -2 \sin(x) \cos(x) $$

This is equal to $ -v(x) $, so the function satisfies the property of odd functions.

Therefore, $ v(x) = 2 \sin(x) \cos(x) $ is an odd function.