How do you find the linearization of $f (x) = cos x$ $f(x)=cosx$ at x=5pi/2?

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Answer 1 · 2016-12-16T13:30:16

$cos x ≅ \frac{5 π}{2} - x$ $cosx~=(5pi)/2-x$

The linear approximation of a differentiable function around $x = ¯ x$ $x=barx$ is given by the tangent line, so that:

$f(x)~=f(barx)+f'(barx)(x-barx)$

For $f(x) = cosx$ at $barx= (5pi)/2$ .

$cosx~=cos((5pi)/2) -sin((5pi)/2)(x-(5pi)/2)$

Considering that:

$(5pi)/2 = 2pi+pi/2$ ,
$cos(x+2pi)= cosx$
$sin(2pi+x) = sinx$
$cos(pi/2)=0$
$sin(pi/2)=1$

This becomes:

$cosx~=(5pi)/2-x$

How do you find the linearization of f(x)=cosxf(x)=cosxf(x)=cosx at x=5pi/2?