Single-Variable-Calculus · Unit 3 · Video 5 · Interactive Practice

A Second, Purely Geometric Proof | The Right Triangle in Circular Motion

IKey Formulas

FormulaNameWhat it says
y=sin⁡θy = \sin\thetaSine as a heightThe height of the point PP at angle θ\theta on the unit circle
Δy=PR=sin⁡(θ+Δθ)−sin⁡θ\Delta y = PR = \sin(\theta + \Delta\theta) - \sin\thetaChange in heightThe vertical side of triangle PQRPQR — not the chord PQPQ
PQ≈arc PQ=1⋅Δθ=ΔθPQ \approx \text{arc } PQ = 1 \cdot \Delta\theta = \Delta\thetaHypotenuseArc length is radius times angle in radians, and the radius is 11
ΔyΔθ=sin⁡(θ+Δθ)−sin⁡θΔθ\dfrac{\Delta y}{\Delta\theta} = \dfrac{\sin(\theta + \Delta\theta) - \sin\theta}{\Delta\theta}Difference quotientIts limit as Δθ→0\Delta\theta \to 0 is ddθsin⁡θ\dfrac{d}{d\theta}\sin\theta

Key Insight: The whole proof happens inside one small right triangle PQRPQR — right angle at RR, vertical side Δy\Delta y, hypotenuse essentially Δθ\Delta\theta — and it works at every θ\theta, not just at θ=0\theta = 0.

IIVisualization 1 — Sine as a Height

The sine of θ\theta is a height: the vertical coordinate of the point at angle θ\theta on the unit circle.

IIIVisualization 2 — Which Segment Is Δy\Delta y?

Moving the particle from PP to QQ changes the height by the vertical side PRPR, not by the chord PQPQ.

IVVisualization 3 — The Angle at PP

One piece of the triangle is still unmeasured: the angle at PP between PRPR and the chord PQPQ.

💡 The next step pins ∠QPR\angle QPR down from where the triangle sits on the circle; right-triangle trigonometry then turns the hypotenuse Δθ\Delta\theta into the vertical side Δy\Delta y.

VQuiz Questions

Problem 1 · Which Segment Is the Change in Height?

Given: PP sits at angle θ\theta on the unit circle, QQ at angle θ+Δθ\theta + \Delta\theta, and RR is the foot of the perpendicular dropped from QQ onto the vertical line through PP. Which length equals sin⁡(θ+Δθ)−sin⁡θ\sin(\theta + \Delta\theta) - \sin\theta?

✅ Correct! PRPR is the only vertical segment in the picture, so its length is exactly the change in height: Δy=PR\Delta y = PR.
❌ Close, but that is the hypotenuse. PQPQ runs from start to finish, so it carries the sideways motion as well as the rise — it is longer than Δy\Delta y.
❌ That is the horizontal side. RQRQ measures the sideways part of the motion, cos⁡(θ+Δθ)−cos⁡θ\cos(\theta+\Delta\theta) - \cos\theta in size, not the change in height.
❌ Not quite. A change in height is measured straight up, so look for the segment that is vertical.
Show solution

Heights are measured from the horizontal axis: PP has height sin⁡θ\sin\theta and QQ has height sin⁡(θ+Δθ)\sin(\theta+\Delta\theta).

Since RR lies on the vertical line through PP, it has the same first coordinate as PP; and since QRQR is horizontal, RR has the same height as QQ. So

R=(cos⁡θ, sin⁡(θ+Δθ)),PR=sin⁡(θ+Δθ)−sin⁡θ=ΔyR = \big(\cos\theta,\ \sin(\theta+\Delta\theta)\big), \qquad PR = \sin(\theta+\Delta\theta) - \sin\theta = \Delta y

The chord PQPQ is the hypotenuse of the right triangle PQRPQR and RQRQ is its horizontal side; only PRPR is the rise.

Problem 2 · Arc Length Wants Radians

Given: PP and QQ lie on the unit circle with central angle ∠POQ=30°\angle POQ = 30\degree — find the length of the arc from PP to QQ.

✅ Correct! 30°=π/630\degree = \pi/6 radians, and on the unit circle the arc is 1⋅Δθ=π/6≈0.5241 \cdot \Delta\theta = \pi/6 \approx 0.524.
❌ Those are degrees, not a length. arc=r Δθ\text{arc} = r\,\Delta\theta holds only when Δθ\Delta\theta is measured in radians — this is exactly why the proof works in radians.
❌ That is a height, not an arc. 0.5=sin⁡30°0.5 = \sin 30\degree measures how high the point sits, while the arc measures how far it travelled.
❌ Not quite. Convert 30°30\degree to radians first, then multiply by the radius 11.
Show solution

Step 1 — convert to radians:

30°=30⋅π180=π630\degree = 30 \cdot \frac{\pi}{180} = \frac{\pi}{6}

Step 2 — arc length is radius times angle:

arc PQ=r Δθ=1⋅π6=π6≈0.524\text{arc } PQ = r\,\Delta\theta = 1 \cdot \frac{\pi}{6} = \frac{\pi}{6} \approx 0.524

The chord is slightly shorter, 2sin⁡15°≈0.5182\sin 15\degree \approx 0.518 — that tiny gap is the entire error in the approximation PQ≈ΔθPQ \approx \Delta\theta, and it shrinks to nothing as Δθ→0\Delta\theta \to 0.

Problem 3 · Reading the Triangle with Trigonometry

Given: in right triangle PQRPQR the right angle is at RR, the hypotenuse is PQ≈ΔθPQ \approx \Delta\theta, and the angle at PP is α=∠QPR\alpha = \angle QPR. For the second part take α=60°\alpha = 60\degree and Δθ=0.02\Delta\theta = 0.02.

First, which expression gives the vertical side PR?

Then, how big is that side in numbers?

✅ Correct! PRPR is the side adjacent to α\alpha, so PR=PQcos⁡α≈Δθcos⁡αPR = PQ\cos\alpha \approx \Delta\theta\cos\alpha; at α=60°\alpha = 60\degree that is half of Δθ\Delta\theta, namely 0.010.01.
❌ That is the other side. sin⁡α\sin\alpha picks out the side opposite α\alpha, which is the horizontal side RQRQ — the sideways motion, not the rise.
❌ Check which side is longest. Dividing by cos⁡α\cos\alpha makes Δy\Delta y larger than the hypotenuse PQPQ, and a leg can never beat the hypotenuse.
❌ Not quite. Label the sides first: PRPR is adjacent to α\alpha, RQRQ is opposite it, and PQPQ is the hypotenuse.
❌ That used the sine. 0.02sin⁡60°=0.01730.02\sin 60\degree = 0.0173 is the horizontal side RQRQ; the rise uses the cosine.
❌ Recompute. cos⁡60°=12\cos 60\degree = \tfrac12, so Δy≈0.02⋅12\Delta y \approx 0.02 \cdot \tfrac12.
Show solution

In a right triangle, the leg adjacent to an angle equals the hypotenuse times the cosine of that angle. Here the right angle is at RR, so PQPQ is the hypotenuse and PRPR sits next to α\alpha:

Δy=PR=PQcos⁡α≈Δθcos⁡α\Delta y = PR = PQ\cos\alpha \approx \Delta\theta\cos\alpha

With α=60°\alpha = 60\degree and Δθ=0.02\Delta\theta = 0.02:

Δy≈0.02⋅cos⁡60°=0.02⋅12=0.01\Delta y \approx 0.02 \cdot \cos 60\degree = 0.02 \cdot \tfrac{1}{2} = 0.01

Dividing by Δθ\Delta\theta leaves the difference quotient ΔyΔθ≈cos⁡α\dfrac{\Delta y}{\Delta\theta} \approx \cos\alpha — which is why finding α\alpha finishes the proof.

Problem 4 · Over the Top of the Circle

Given: the particle is at the top of the unit circle, θ=90°\theta = 90\degree, and sweeps a small Δθ\Delta\theta onward from there — compare the change in height Δy\Delta y with Δθ\Delta\theta.

✅ Correct! At the top the particle is moving horizontally, so the chord is nearly all sideways: ∠QPR≈90°\angle QPR \approx 90\degree and Δy≈Δθcos⁡90°=0\Delta y \approx \Delta\theta\cos 90\degree = 0.
❌ The height does fall — but far more slowly than that. Exactly, Δy=cos⁡Δθ−1≈−12(Δθ)2\Delta y = \cos\Delta\theta - 1 \approx -\tfrac{1}{2}(\Delta\theta)^2, which for Δθ=0.02\Delta\theta = 0.02 is only −0.0002-0.0002.
❌ Not quite. Picture the motion at the very top of the circle: which way is the particle heading, and how much of that is vertical?
Show solution

Compute the change in height directly at θ=π2\theta = \tfrac{\pi}{2}:

Δy=sin⁡ ⁣(π2+Δθ)−sin⁡π2=cos⁡Δθ−1≈−12(Δθ)2\Delta y = \sin\!\left(\tfrac{\pi}{2} + \Delta\theta\right) - \sin\tfrac{\pi}{2} = \cos\Delta\theta - 1 \approx -\tfrac{1}{2}(\Delta\theta)^2

So Δy\Delta y is second order in Δθ\Delta\theta: for Δθ=0.02\Delta\theta = 0.02 the height changes by −0.0002-0.0002, one hundredth of Δθ\Delta\theta — essentially zero next to a first-order quantity.

The triangle says the same thing. At the top the chord PQPQ is nearly horizontal, so the angle at PP is close to 90°90\degree and the vertical leg is PR≈Δθcos⁡90°=0PR \approx \Delta\theta\cos 90\degree = 0. This is the geometric face of cos⁡π2=0\cos\tfrac{\pi}{2} = 0: the derivative of sin⁡θ\sin\theta vanishes at the top of the circle.

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