How do you find the linearization at a=pi/4 of #f(x)=cos^2(x)#?

Question

Emilia Aguilar · Accepted Answer

#y=1/2+(pi/4-x)# A linearization is typically in the form: #y=f(a)+f'(a) (x-a) # where #a=pi/4# Finding #f(a)# #f(pi/4)=cos^2(pi/4)=(sqrt(2)/2)^2=1/2# Finding #f'(a) # #f'(a) =-2sin^2(pi/4)=-2(sqrt(2)/2)^2=-1# Putting it all together: #y=1/2+(pi/4-x)#

Isaiah Allen · Answer

undefined To find the linearization at $a = \frac{\pi}{4}$ of $f(x) = \cos^2(x)$, follow these steps:

1. Find the first derivative of $f(x)$, $f'(x)$.
2. Evaluate $f'(a)$ at $a = \frac{\pi}{4}$.
3. Use the formula for linearization: 
$$L(x) = f(a) + f'(a)(x - a)$$
where $L(x)$ is the linearization of $f(x)$ at $a$, $f(a)$ is the value of $f(x)$ at $a$, and $f'(a)$ is the derivative of $f(x)$ at $a$.
4. Plug in the values you found into the linearization formula to get the linearization at $a = \frac{\pi}{4}$.

Let's solve it step by step:

1. $f(x) = \cos^2(x)$
   $f'(x) = -2\cos(x)\sin(x)$
  
2. Evaluate $f'(\frac{\pi}{4})$:
   $f'(\frac{\pi}{4}) = -2\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}) = -\sqrt{2}$
  
3. Use the linearization formula:
   $L(x) = f(\frac{\pi}{4}) + f'(\frac{\pi}{4})(x - \frac{\pi}{4})$
  
4. Plug in the values:
   $L(x) = \cos^2(\frac{\pi}{4}) - \sqrt{2}\left(x - \frac{\pi}{4}\right)$
   Simplify as needed.