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

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Answer 1 · 2018-04-13T01:50:37

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

The linearization at #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.