## Calculus: Early Transcendentals 8th Edition

$f(x) = \frac{\sqrt 3}{2}\times(x- \pi/6) + 0.5$
To establish a linearization, you must first find $f(a)$ and $f'(a)$ and make it into the following form:$f(x)=f(a)+f'(a)\times(x-a)$ . a = $\pi/6$ in this particular case. $f'(a)$ = cos($\pi/6$) or $\sqrt 3 \div 2$ $f(a)$ = sin($\pi/6$) or 1/2