# Find linearization of f(x)=sinx at x=(pi/6)? Use your linearization to approximate sin(pi/3)?

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.
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