For what values of $k$ is $f(x)=\cos(x-k)$ even, odd or neither

Question

I am asked the following question

For what values of $k$ is $f(x)=\cos(x-k)$

a) even

b) odd

c) neither

I am not quite sure how to tackle this problem, I only got so far as to say, for example in letter a,

$$f(-x) = f(x)$$ $$\cos(-x-k) = \cos(x-k)$$

I'm not sure how to continue from here.

Answer 1

The easy way to do this is look at the plot, and think about the places you might be able to drop a vertical line such that the function reflects across the line.

Once you have you points, can you find a rigorous justification.

Another way you could do it would be to use the angle additions rules

$cos(\theta - k) = \cos \theta\sin k + \sin\theta\sin k$

We know that $\cos \theta$ is even, and $\sin \theta$ is not. Is there a value of $k$ such that the $\sin \theta$ term drops away?

Answer 2

Hint:

Use Prosthaphaeresis Formulas

For the even case,

$$\cos(-x-k)=\cos(x-k)\iff2\sin x\sin k=0$$

As $x$ is arbitrary, $\cos x\ne0\implies\sin k=0$

Similarly for the odd case