The first second-order equation on this course is solved by two functions you already know, and the reason is a fact about their rates: the rate of sin is cos, and the rate of cos is −sin. The quotients that show it cannot be fractions, so the lab rounds them, labels them with ≈, and prints the number they head for. Taking a rate twice then returns each function with its sign reversed, which is exactly what the next lesson's equation asks for.
Core ideas
Three ideas. The first two are the facts the course leans on; the third is what they give when used twice, and it is the whole reason for the lesson.
The quotient (f(a + h) − f(a))/h is the average rate over a step of width h. When it settles toward one number as h is halved again and again, that number is the rate at a. For sin and cos the quotients are irrational, so the lab rounds each to six significant figures and puts ≈ in front of every one; no entry in these tables is exact.
At a = 0 the quotients of sin head for 1, which is cos 0. At a = 0 the quotients of cos head for 0, which is −sin 0, and at a = 1 they head for a negative number. The pattern across every place tried is sin′ = cos and cos′ = −sin, with t measured in radians, and the lesson states it as a claim the table demonstrates.
If cos′ = −sin and sin′ = cos, then cos″ = (−sin)′ = −cos and sin″ = (cos)′ = −sin. Each function is the negative of its own second rate. That is a differential equation in one line, y″ = −y, and both functions satisfy it.
The material
The claim, how far a rounded table can carry it, one case where the sign is easy to lose, and the step to the second rate.
Every earlier rate in this Subject came out as a fraction, because the functions were polynomials and the quotient was a polynomial in h. sin and cos are not polynomials, and at almost every place the value is irrational, so no column of fractions is available. The lab here computes the quotients in ordinary floating point, prints each rounded to six significant figures behind ≈, and prints the number they head for. The honest status of that last number is the important point, and it is stated below. The lab's fourth tile divides the last quotient by f(a); it was built for the exponential, whose rate is a fixed multiple of its value, and for sine and cosine it says nothing, so the first three tiles are the ones to read.
For t measured in radians, the rate of sin at every t is cos t, and the rate of cos at every t is −sin t. In symbols: sin′ t = cos t and cos′ t = −sin t.
This is a claim, and the lab demonstrates it at the places you choose without proving it. A table of quotients that settle toward a number shows the pattern; it cannot show that the pattern continues for every smaller h, and no table can. The course uses the claim from here on, as every course in this Subject uses the rate of an exponential, and says so once.
At a = 0 the first sine preset starts with h = 1. The quotient is (sin 1 − sin 0)/1 = sin 1, which is ≈ 0.841471 and has no fraction form. Halving h gives ≈ 0.958851, ≈ 0.989616 and ≈ 0.997398, and the lab continues to the sixth halving. The numbers climb toward 1, which is cos 0, and the gap to 1 is cut to about a quarter each time h is halved. A quarter, and not the half that the cosine table below will show, because the leading error of a one-sided quotient is proportional to h times the second rate of the function at a, and sin″ 0 = −sin 0 = 0: at 0 that term is absent and the next one, proportional to h², is what remains.
The cosine preset at a = 0 is the one where the sign hides. Its quotients start at ≈ −0.459698 and shrink toward 0, because cos is at the top of its hill at 0 and is not yet falling. That does not tell you the sign of cos′ in general, since 0 has no sign. The preset at a = 1 does: its quotients are all negative and head for −sin 1 ≈ −0.841471, because the cosine is falling at 1. A reader who remembers only that the rate of cos is “the other one” will write sin 1, and the table refutes it with a minus sign.
The first quotient, h = 1, is cos 2 − cos 1, which the lab prints rounded as ≈ −0.956449. It is negative because cos is smaller at 2 than at 1.
The last of the seven quotients, at h = 1/64, is ≈ −0.845658. It is still moving toward the limit −sin 1 ≈ −0.841471, and from the third row on the gap shrinks by about a half each time, which is what a quotient with an error proportional to h does.
Radians matter. A degree is π/180 of a radian, so a table built on degrees would head for cos times π/180 instead, and the rates would not be each other's. Everything here, and the whole course, measures the argument in radians.
cos″ t = −cos t and sin″ t = −sin t for every t.
Use the two claims above. The rate of cos is −sin, so the rate of the rate of cos is the rate of −sin, which is −(sin′), and sin′ = cos. Hence cos″ = −cos.
The same steps give sin″ = (cos)′ = −sin. The proof is only as firm as the two claims it uses, which the table demonstrates and does not prove.
□
Both identities say that the second rate of the function is minus the function. Written as an equation for an unknown y, that is y″ = −y, or y″ + y = 0. The next lesson substitutes into it and finds that cos, sin and every combination of them solve it.
Interactive
Rounded quotients, labelled, and what they approach
Method
Five moves. The last is the one that makes the lesson: a rate, used twice.
Pick sin or cos and a place a. Write down, before reading the table, what you expect the rate to be at that place, with its sign.
The first is the average rate over the widest step, the last over the narrowest. Both are rounded, and the tile says so with ≈.
Each halving of h should move the quotient closer to a single number. If the changes are not shrinking, the table is not showing a rate.
The lab prints the number the quotients head for, as cos(1) ≈ 0.540302 for example. That is the claim, rounded, and the last quotient is near it but not equal to it.
To get a second rate, take the rate of the result: cos becomes −sin, which becomes −cos. Count the minus signs as you go.
Read four quotients of sin at 0, state what they head for, and use the two claims to find cos″.
Every entry in the column is rounded, and the lab labels it so; the limit 1 is stated as a claim and used from here on.
The second line of the last pair does the work of the lesson: the minus sign in cos′ = −sin stays when the rate is taken again, so cos″ is −cos and not cos.
You should be able to say what the quotients of sin and cos head for at a place you choose, state sin′ = cos and cos′ = −sin as claims the table demonstrates, say which entries are rounded, and conclude that the second rate of each is the negative of the function.
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