# Is cos(x^{3} - x + 2) even or odd?

Trigonometry is the branch of maths that deals with the relationships between sides and angles of a right angles triangle.

## Answer: cos (x^{3} - x + 2) is neither even nor odd.

Let's understand the solution.

**Explanation:**

To check for an odd function f(x), we need to verify if f(-x) = -f(x), and to check for even functions we check if it follows the relation f(-x) = f(x).

Now, we check this for cos (x^{3} - x + 2):

⇒ f(x) = cos (x^{3} - x + 2)

Since cos x is positive on the first and the fourth quadrant and -x is the angle taken from the x-axis in a clockwise direction, hence lying in the fourth quadrant. We have cos (-x) = cos x.

Now further:

⇒ f(-x) = cos ((-x)^{3} - (-x) + 2)

⇒ f(-x) = cos (-x^{3} + x + 2) ≠ cos (x^{3} - x + 2)

⇒ - f(x) = - cos (x^{3} - x - 2) ≠ cos (x^{3} - x + 2)

⇒ f(-x) ≠ - f(x) and f(-x) ≠ f(x)

Hence, it neither follows conditions for odd functions nor follows for even functions.