Numbers & Functions · 6 Trigonometric Functions

6.10 The surprise, armed (but not yet sprung)

Think about how strange it is that sine and cosine have those addition formulas at all. We just proved them, so we know they’re true. But the proof was fiddly — it tells you the formulas are correct without telling you why they have that particular shape. It looks almost like a coincidence. It is not. There’s a hidden machine underneath sine and cosine, and when we meet it in Chapter 8 the addition formulas will be the obvious, one-line consequence of this: eiθ = cos θ + i sin θ.

You can’t read that yet — the i is the new number we haven’t built, and why e (the growth number!) would have anything to do with circles is a genuine shock. But there it is, the most famous equation in mathematics, sitting and waiting. Hold the thread.

6.11 Where we are now

We met a completely new kind of machine — bounded and periodic, the mathematics of everything that repeats. We built sine and cosine from one honest picture: the walker on the unit circle. We saw why π governs angles, why the sine wave is the master shape of every oscillation, and how tangent is their Trigonometric Functions ratio. We found the Pythagorean identity — Chapter 2’s Pythagoras standing on a circle — and a small, connected toolkit, with a promise that the fearsome addition formulas are about to collapse into one beautiful line. Things worth thinking about 1. Read off the circle, no chart: where is the walker when sin θ = 1? When cos θ = −1? Can sin θ and cos θ both equal 1 at once? (Use the Pythagorean identity — if both were 1, what would cos2 + sin2 come to?)

2. The sine wave repeats every 2π. If you wanted a wave that repeated every 1 unit instead, what could you do to the input of sin θ to speed the walker up? (Think about sin(2πθ).)

3. Tangent jams wherever cos θ = 0. Describe, in terms of the walker and the line from the center, why the slope becomes impossible to write down at exactly those spots. What is the line doing there?

4. (Looking ahead.) We claimed the growth number e is secretly connected to circles. On its face that’s absurd. What would have to be true about the new numbers of Chapter 8 for a machine that explodes and a machine that loops to turn out to be the same machine?

Problems to work Problem 1 — Read the circle.

Without any chart, give sin θ and cos θ for each, saying where the walker stands. (a) θ = 0; (b) π 2 ; (c) π; (d) 3π 2 ; (e) 2π (what do you notice versus θ = 0?).

Problem 2 — Use the Pythagorean identity.

A walker has cos θ = 3 5. Using cos2 θ + sin2 θ = 1, find the two possible values of sin θ. (Why two? One walker is above the axis, one below.)

Problem 3 — Derive a formula yourself.

Set α = β = θ in sin(α + β) = sin α cos β + cos α sin β to derive sin(2θ). (You should get 2 sin θ cos θ in one line — it was hiding inside the addition formula all along.)

Problem 4 — Derive the other double-angle, three ways.

Set α = β = θ in the cosine addition formula to get cos(2θ) = cos2 θ −sin2 θ. Then use the Pythagorean identity to rewrite it using only sin θ, and again using only cos θ. (You’ll find cos 2θ = 1 −2 sin2 θ and cos 2θ = 2 cos2 θ −1 — three faces of one fact.)

Problem 5 — Recover the special angles from the triangles.

Without looking back, draw the two triangles and rebuild them. (a) From half a unit square, find sin π 4 and cos π 4 . (b) From half an equilateral triangle, find sin π 3 , cos π 3 , sin π 6 , cos π 6 . If you can redraw the triangles, you own the values.

Problem 6 — Read reductions off the walker.

Use the walker picture (not memorized formulas): (a) sin 2π 3 (hint: 2π 3 = π −π 3 , a reflection across the vertical); (b) cos 2π 3 ; (c) cos π 3 using that it equals sin π 6 (the complement swap).

Problem 7 — Derive tan(2θ).

From tan(2θ) = 2 sin θ cos θ cos2 θ−sin2 θ, divide top and bottom by cos2 θ and simplify, using tan θ = sin θ cos θ, to show tan(2θ) = 2 tan θ 1−tan2 θ. (A genuine power tool for the calculus book.)

⋆Teach it — explain it to someone Problem 8 — Teach it. Explain to someone why sin θ can never be bigger than 1 or smaller than −1, using the walker-on-a-circle picture — not a formula. The cleanest explanation is purely visual: the walker is stuck on a circle of radius 1, so how high can they possibly get? If your listener sees why a bounded machine is fundamentally different from the exponential that ran off to infinity, you’ve understood what makes these functions special.

Going further ⋆Teach it — explain it to someone Problem 9 — Teach it (why cos is “co-sine”). Using only the unit circle, show a family member why sin(π 2 −θ) = cos θ. Draw the walker at θ, then at its mirror image across the 45◦diagonal, and point out that reflecting swaps “how high” with “how far across.” The punchline to land: cosine isn’t a separate function to fear — it’s just sine measured from the perpendicular direction. That’s why the two waves look identical but shifted.

Trigonometric Functions Problem 10 — State the domains.

For each function, say which angles are forbidden, using “a trig function jams where its denominator is zero” and the walker picture. (a) tan θ; (b) cot θ; (c) csc θ = 1 sin θ (same jams as which other function?); (d) why do sin θ and cos θ have no forbidden angles?

Problem 11 — [challenge] Build a wave to order.

You want a wave that oscillates between −3 and 3 (instead of −1 and 1) and repeats every π units (instead of 2π). Starting from sin θ, write the formula. (Hint: one number out front controls the height; one number inside, multiplying θ, controls how fast the walker goes around — and you met “inside affects the horizontal” in the transformation preview. Which inside-multiplier halves the period?)

Problem 12 — [challenge] A surprising sum.

Use the addition formulas to compute sin(75◦) exactly, without a calculator. (Hint: 75◦= 45◦+ 30◦, and you know the sine and cosine of both from the two triangles. Plug into sin(α + β) and simplify to a single expression with √ 2 and √ 3.) Check that your exact answer is about 0.966.

Problem 13 — [challenge] The identity behind a sunset.

Prove the identity sin2 θ = 1−cos 2θ 2 starting from the double-angle formula cos 2θ = 1 −2 sin2 θ and just solving for sin2 θ. (This innocent-looking rearrangement is one of the most-used tricks in all of calculus — it’s how you integrate sin2, which shows up everywhere from the energy in a sound wave to the blue of the sky.)

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