# Chapter 7: Trig Limits and Derivatives

> One geometric limit — sin θ over θ — powers every trig derivative in the syllabus, from sin x to sin(x²) to the acceleration a = −ω²x of anything oscillating.

**Canonical:** https://calculus.kitzo.work/chapter/trig-derivatives  
**Level:** Core  
**Estimated time:** 40 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- lim sin x/x = 1
- The small case
- The large case
- The other case
- Derivatives of sin & cos
- Simple harmonic motion

**Why this matters**
Differentiating sin and cos is impossible from the power rule alone — it rests on one geometric fact: sin θ ≈ θ for tiny θ. Lock that limit in and the whole trig toolkit unlocks: derivatives, chain-rule composites, and the equation of motion of every spring, pendulum and sound wave you will ever model.

## 1 · The one fact to know: sin θ / θ → 1

On a unit circle, take a small angle θ. The vertical chord has height sin θ, the arc has length θ, and the tangent segment has length tan θ. Geometry gives the squeeze sin θ < θ < tan θ, so dividing by sin θ yields 1 < θ/sin θ < 1/cos θ, which rearranges to cos θ < sin θ / θ < 1. As θ → 0, cos θ → 1, so by the squeeze theorem sin θ / θ is trapped between 1 and 1 and must approach 1. Every line of this section rests on angles measured in radians — in degrees the same limit equals π/180.

## 2 · The small case: θ → 0 in disguise

With the key fact in hand, a family of limits collapses instantly. Rewrite sin(kx)/x as k · sin(kx)/(kx), which tends to k. Hence sin 3x/x → 3, sin 5x/sin 2x → 5/2, and x/sin x → 1. For the cos version, use 1 − cos x = 2 sin²(x/2): then (1 − cos x)/x² = ½ · (sin(x/2)/(x/2))² → ½. Memorise that pair: sin x/x → 1 at zero, but (1 − cos x)/x² → ½.

## 3 · The large case: x → ∞

Sine never escapes its bounds: |sin x| ≤ 1 for every x. Therefore |sin x / x| ≤ 1/|x|, which shrinks to 0 as x → ∞ (also as x → −∞). By the squeeze theorem, lim x→∞ sin x / x = 0 — no matter how wildly sin x oscillates on top, division by an unbounded x flattens it. The same holds for cos x / x.

## 4 · The other case: 1 − cos x

The remaining relative, lim x→0 (1 − cos x)/x, equals 0: write (1 − cos x) = 2 sin²(x/2) and note the numerator is quadratic in x while the denominator is linear, so the quotient vanishes. Equivalently, (cos x − 1)/x → 0 and (cos x − 1)/x² → −½. When you see a cosine difference over a power of x, decide whether x or x² is doing the dividing — that decides between 0 and the half.

**Worked example 7.1 — Three limits, one toolkit**
**(a)** lim x→0 sin 6x / x = lim x→0 6 · sin 6x/6x = 6·1 = 6.
**(b)** lim x→0 (1 − cos 2x)/x². Since 1 − cos 2x = 2 sin²x, this equals 2·lim (sin x/x)² = 2·1 = 2.
**(c)** lim x→∞ (sin 3x)/x = 0, since |sin 3x| ≤ 1 and 1/x → 0.
**Step habit:** always bend the expression into the shape sin(□)/(□) or (1 − cos □)/□², then quote the memorised value — never the other way round.

## 5 · Where (sin x)′ = cos x and (cos x)′ = −sin x come from

Run the difference quotient on sin x: (sin(x+h) − sin x)/h = sin x·(cos h − 1)/h + cos x·(sin h)/h, using the angle-addition formula sin(x+h) = sin x cos h + cos x sin h. The section-1 limits send (cos h − 1)/h → 0 and (sin h)/h → 1, leaving cos x. Repeating with cos x gives −sin x — the minus sign is not decoration, it falls straight out of the cos angle-addition formula.

## 6 · The chain rule with trig

Composite trig functions differentiate outside-in exactly as before: d/dx sin(x²) = cos(x²)·2x, d/dx cos(3x + 1) = −3 sin(3x + 1), d/dx sin²x = 2 sin x cos x. Watch the inner derivative accumulate a factor every single time — cos(2x) missing its 2 is the classic error.

## 7 · Simple harmonic motion

Whenever acceleration is proportional to negative displacement, a = −ω²x with ω > 0, the motion is simple harmonic: x = A cos(ωt + φ), where A is amplitude, ω is angular frequency, and φ is phase. Two derivatives bring you back to a = −Aω²cos(ωt + φ) = −ω²x, consistent with the equation. The period is T = 2π/ω — four ω's per second means T = π/2 ≈ 1.57 s, and maximum speed is Aω at the equilibrium point. A mass on a spring and a small-angle pendulum are the canonical examples.

**Watch out**
All of these limits and derivatives are legal only in radians. If a problem sneaks in degrees, convert first — differentiating sin(x°) introduces a factor π/180 at every step, and the limit sin θ/θ → 1 becomes π/180.

**Chapter formula sheet**
lim x→0 sin x / x = 1
lim x→∞ sin x / x = 0
lim x→0 (1 − cos x)/x = 0,  lim x→0 (1 − cos x)/x² = ½
(sin x)′ = cos x,  (cos x)′ = −sin x
d/dx sin(u) = u′ cos u,  d/dx cos(u) = −u′ sin u
SHM: a = −ω²x,  x = A cos(ωt + φ)
T = 2π/ω,  v_max = Aω

## Common mistakes

- Quote sin θ/θ → 1 in degrees — the answer silently becomes π/180.

- Writing (cos x)′ = sin x — the sign is negative: (cos x)′ = −sin x.

- Dropping the inner factor: d/dx sin(x²) = cos(x²)·2x, not cos(x²).

- Confusing the large case with the small case — as x → ∞ the ratio goes to 0, never to 1.

- Reaching for (1 − cos x)/x² → 1 and stopping — it converges to ½, so keep the ½.

- Computing SHM period as ω instead of 2π/ω, or quoting Aω as the maximum acceleration (it is Aω²).

## Practice exercises

1. Evaluate lim x→0 sin(5x)/x.

2. Evaluate lim x→0 (1 − cos x)/x².

3. Evaluate lim x→0 sin(7x)/sin(3x). Give your answer as a decimal to two decimals.

4. Differentiate y = sin(x²). Give your answer in terms of x.

5. d/dx(cos x) equals:

- A. sin x
- B. −sin x
- C. cos x
- D. −cos x

6. A particle has displacement x = 4 cos(3t) (metres). What is its acceleration (in m/s²) at t = 0?

7. Differentiate y = cos(3x).

8. Which equation defines simple harmonic motion?

- A. a = ω²x
- B. a = −ω²x
- C. a = −ω²/x
- D. a = ω²/x

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