How to Find the Linearization of $f(x) = 8\cos^2(x)$ at the Point Where $x = \pi/4$ I am looking to find the linearization of $f(x) = 8\cos^2(x)$ at the point where $x = \pi/4$
Now, I know that the general formula for finding linearization is $L(x) = f(x) + f'(x) (x-a)$. So, I have made the following steps to get my answer:
1) Use the above formula to get $L(x) = 8\cos^2(\pi/4)+16\cos(\pi/4)(-\sin(\pi/4))(\pi/4-a)$
2) Simplify to get $4-8(\pi/4-a)$
However, I am not sure if this is completely correct.
Does anyone know whether this is the correct way to find this answer, and if it is not, how to arrive at the true answer?

here is a way to check your answer. $y = 8\cos^2 x = 4 + 4 \cos(2x).$  making a change of variable $x = \pi/4 + h,$ we have $y = 4 + 4\cos(\pi/2 + 2h)=4 - 4\sin(2h) = 4-8h+\cdots$  for $h = 0+ \cdots$ 
that is $$y = 8\cos^2 x = 4 - 8(x - \pi/4) + \cdots \text{ for } x = \pi/4 + \cdots $$
