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

The linearization is the tangent line. So

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To find the linearization of ( f(x) = \sin(x) ) at ( a = \frac{\pi}{6} ), follow these steps:

- Compute the first derivative of ( f(x) = \sin(x) ), which is ( f'(x) = \cos(x) ).
- Evaluate ( f'(a) ), where ( a = \frac{\pi}{6} ). So, ( f'\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} ).
- The linearization of ( f(x) ) at ( a ) is given by the equation ( L(x) = f(a) + f'(a)(x - a) ).
- Substitute ( f(a) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} ) and ( f'(a) = \frac{\sqrt{3}}{2} ) into the equation to get ( L(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) ).

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