**Answer 1**

Linearization:

f(x) = sin(x)

L(x) = f(a) + f'(a)(x-a)

f'(x) = cos(x)

a = π/6

L(x) = sin(π/6) + cos(π/6)(x-π/6)

L(x) = 1/2 + (√3/2)(x – π/6)

L(x) = 1/2 + (√3/2)x – π√3/12

L(x) = (6-π√3)/12 + (√3/2)x

or

L(x) = 0.04655 + 0.86603x

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Approximation:

x = π/3

sin(π/3) = 0.86603

L(π/3) = (6-π√3)/12 + (√3/2)(π/3)

L(π/3) = (6 – π√3 + 2π√3)/12

L(π/3) = (6+π√3)/12 ≈ 0.95345

The linearized approximation is (0.95345/0.86603 – 1) = 10% too high.

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Try an approximation at x closer to the linearization point:

Approximate sin(x) at x = π/6 + 0.1 = 0.62360

sin(π/6 + 0.1) = sin(0.62360) = 0.58396

L(0.62360) = 0.04655 + 0.86603(0.62360) = 0.58661

The linearized approximation is (0.58661/0.58396 – 1) = 0.5% too high

The closer x is to the linearization point, x = π/3, the closer the approximated value is to the true value.

**Answer 2**

The linear approx of f(x) at a is

L(x) = f(a) + f'(a)(x-a)

= sin(pi/6) + cos(pi/6)(x-pi/6) ……. f'(x) = cos(x)

= 1/2 + sqrt(3)/2(x-pi/6)

L(pi/3) = 1/2 + sqrt(3)/2 *(pi/3-pi/6)

= 1/2 + sqrt(3)/2 *pi/6