Comparing Linear Approximations to Calculator Computations In lecture, we explored linear approximations to common functions at the point x = 0. In this worked example, we use the approximations to calculate values of the sine function near x = 0 and compare the answers to those on a scientiﬁc calculator. Find the linear approximation to sin(x) at the point x = 0 and use your answer to approximate the values of sin(.01), sin(.1) and sin(1). Check your answer on a calculator. Solution: Recall that linear approximation to a function f (x) at a point x = a just means that we use the tangent line T (x) to f (x) at x = a to approximate the function. In the case where f (x) = sin(x), we saw from lecture that the tangent line T (x) at x = 0 was given by: T (x) = f � (0)(x − 0) + f (0) = cos(0)(x − 0) + sin(0) = x. In short, we write sin(x) ≈ x when x ≈ 0. (Try drawing a picture of the sine curve and its tangent line at x = 0 to illustrate this.) So we would approximate the values of sine above as follows: sin(.01) ≈ .01 sin(.1) ≈ .1 sin(1) ≈ 1 Our expectation is that the closer we choose our estimation point to x = 0 (where the tangent line meets the function), the better our approximation. And indeed the calculator conﬁrms: sin(.01) = .00999983333... sin(.1) = .099883341... sin(1) = 0.84147098... where we’ve only recorded the ﬁrst few digits of the decimal expansion. (Be careful to set your caculator to “radians” not “degrees” in computing these.) So sin(.01) diﬀers from our approximation .01 by less than .0000002, a very accurate approximation! On the other hand sin(1) diﬀers from 1 by more than .15. Note the approximation must get worse and worse in our example, as sin(x) is always bounded between −1 and 1, while x continues to grow without bound. What if we wanted to approximate sin(1000)? We’ve seen that the linear approximation at x = 0 would give the answer as 1000. Clearly this is far too big. Instead we should take the tangent line at a point closer to x = 1000 where we know the value of the sine function. The general answer for the tangent line to f (x) = sin(x) at x = a is: y = f � (a)(x − a) + f (a) = cos(a)(x − a) + sin(a) so we should choose a for which we know the values of cos(a) and sin(a). We might choose a = 318π, which is around 999. Alternatively, we could use the fact that sine is periodic, with period 2π, so that sin(1000) = sin(1000 − 318π), and then use the approximation at x = 0 for the latter value sin(1000 − 318π) = sin(.97353...). We can do better still if we use fractional multiples of π and apply trigonometric identities.