Particle acceleration position problem (Calculus)

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A particle's position is given by $x(t) = \cos^2(t) + t$, where time is measured in seconds. Find the particle's approximate position after 6 seconds have passed.

So I find the second derivative, which is:

2(sin^2(t)-cos^2(t))

But the problem tells me: Do not compute exactly here. You should be able to use calculus to get a quick estimate.

So would I use approximation to get this value? How should I approach this?

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Consider that $\cos(6) = \cos(2\pi-6)$ and that $\cos^2(t)\approx \left(1-\frac{t^2}{2}\right)^2\approx 1-t^2$ for small $t$ using the Taylor series approximation of $\cos$. $2\pi-6\approx 0.3$, so $\cos^2(6)\approx 1-0.3^2\approx 0.9$. Therefore, $x(6)\approx 6.9$.