How do you find the linearization at a=pi/6 of $f(x)=sinx$ ?

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Answer 1 · 2017-03-26T12:20:54

$L(x) = 1/2+sqrt3/2(x-pi/6)$

$f(x) = sinx$ , so $f'(x) = cosx$

At $a = pi/6$ , we have $y = f(pi/6) = 1/2$ and $f'(pi/6) = sqrt3/2$ .

The linearization is the tangent line. So

$L(x) = f(a)+f'(a)(x-a) = 1/2+sqrt3/2(x-pi/6)$

How do you find the linearization at a=pi/6 of f(x)=sinxf(x)=sinx?