You want to pick a value of cos(x) where x is near 2π/13, and you know what it is.
you know cos(π/6) = √3/2. 2π/13 = π/6 - π/156, which is pretty close to π/6.
cos(2π/13) ≈ cos(π/6)+(-sin(π/6))(π/156)
= √3/2 - (1/2)(π/156)
= 0.8560
the true value is 0.8855
That seems kind of far off, but remember that at π/6 the curve is fairly sharply curved, so the linear approximation will not be too close.
Approximate cos (2π/13) by using a linear approximation with
f (x) =cos x.
cos (2π/13)≈ f(a) + f ' (a)·h = c,
what is h, a, and c? I'm having problems with this one please help
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