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

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Answer 1 · 2016-12-16T13:30:16

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

The linear approximation of a differentiable function around #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#