How do you find the linearization of $F (x) = cos (x)$ $F(x) = cos(x)$ at a=pi/4?

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Answer 1 · 2018-04-13T01:50:37

$L (x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} (x - \frac{π}{4})$ $L(x)=sqrt2/2-sqrt2/2(x-pi/4)$

The linearization at $x = a$ $x=a$ is given by

$L(x)=f(a)+f'(a)(x-a)$

Knowing $f(x)=cosx, a=pi/4,$ then

$f(pi/4)=cos(pi/4)=sqrt2/2$

$f'(x)=-sinx, f'(pi/4)=-sin(pi/4)=-sqrt2/2$

Our linearization is then

$L(x)=sqrt2/2-sqrt2/2(x-pi/4)$

Further simplification would not necessarily result in a cleaner expression.

How do you find the linearization of F(x) = cos(x)F(x)=cos(x)F(x) = cos(x) at a=pi/4?