Introduction

At the end of the previous post I promised to look next at why differentiating \(\sin x\) makes \(\cos x\) pop out. That is today's story.

Let us start with what is strange about it. The sine curve and the cosine curve look nothing alike. Sine starts at 0 at the origin and rises; cosine starts at its peak of 1 and comes down. Yet differentiate sine — that is, measure the slope at each point of the sine curve and plot a new curve — and out comes cosine.

$$\frac{d}{dx}\sin x = \cos x$$

Why cosine, of all things? Is it a coincidental resemblance? This post builds the answer from a single picture: the two shadows of a point going around a circle. Once the picture stands, we will see that pinning it down as a formula needs exactly one part — the fact that "for a very small angle, the arc and the straight line are indistinguishable."

1. The rotating point — height is sine, the speed of the height is cosine

One convention first. Here the angle \(x\) is measured not in degrees but in radians. A radian measures an angle by "how many arcs of one-radius length" it spans (more here), but what matters right now is one thing: measured in radians, on a circle of radius 1 the size of the angle equals the length of the arc it sweeps. This becomes the key engine later.

Now let a point go around a circle of radius 1 (the unit circle) at a steady speed. After turning through an angle \(x\), its position is this.

$$P = (\cos x,\ \sin x)$$

The horizontal coordinate is cosine, the vertical coordinate (height) is sine. What we want is "the speed at which the height \(\sin x\) changes," that is, \(\dfrac{d}{dx}\sin x\).

  • Intuition: as the point circles, the direction it moves at any instant is always perpendicular to the radius (the tangent direction of the circle). And the speed is a steady 1. So the point's velocity is 'the position arrow turned by 90°.'
  • Explanation: when the position is \((\cos x, \sin x)\), turning it 90° gives the velocity arrow \((-\sin x, \cos x)\). But what we want is the speed at which the height (vertical) changes — that is, the vertical component of the velocity arrow.
  • Formula: the vertical component of the velocity is \(\cos x\).
$$\frac{d}{dx}\sin x = \cos x$$

This is what "two shadows" means. Sine and cosine are not two unrelated strangers; they are the shadow of the same rotation seen vertically (sine) and horizontally (cosine). Measure how fast the vertical shadow rises and falls, and that value matches the position of the horizontal shadow — cosine. When the point passes the bottom or top (angles \(x=\pm\pi/2\)), the height momentarily stops, and at that instant cosine is also 0. When the point passes the right end (\(x=0\)), the height rises fastest, and there cosine is at its maximum of 1. The speed and the cosine march in step everywhere.

2. Pinning it down as a formula — plugging into the definition needs two parts

The picture is persuasive, but let us confirm it by plugging directly into the definition of the derivative. The definition is this: take the difference in height of two points a tiny gap \(h\) apart, divide by \(h\), and send \(h\) to 0.

$$\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h}$$

Here we expand \(\sin(x+h)\) with the angle addition formula.

$$\sin(x+h) = \sin x\cos h + \cos x\sin h$$

Substituting this and grouping by \(\sin x\), it splits into two chunks.

$$\frac{\sin(x+h)-\sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}$$

Now sending \(h\to 0\), the result hinges on two small limits.

$$A = \lim_{h\to 0}\frac{\sin h}{h}, \qquad B = \lim_{h\to 0}\frac{\cos h - 1}{h}$$

If \(A=1\) and \(B=0\), the formula above tidies up cleanly to \(\sin x\cdot 0 + \cos x\cdot 1 = \cos x\). So the fact that the derivative of sine is cosine hinges entirely on these two limits. And as we will see shortly, even the second, \(B=0\), follows from the first, \(A=1\). In the end this whole post stands on a single engine — \(\sin h / h \to 1\).

3. The one engine — for a very small angle, arc and straight line become the same

Why does \(\dfrac{\sin h}{h}\) become 1 as \(h\to 0\)? This is where the radians we agreed on earlier shine.

On the unit circle, for an angle \(h\) (in radians), we can compare three lengths.

  • The arc length — by the radian convention, exactly \(h\).
  • \(\sin h\) — the height (vertical shadow) of the point that angle makes.

When the angle is large, these two differ quite a bit. The arc bends around the circle, while \(\sin h\) is a straight line dropping vertically. But what happens as we shrink the angle?

  • Intuition: zoom into a very small sector, and the once-curved arc flattens into an almost straight line. A curve, cut short enough, is indistinguishable from a straight line.
  • Explanation: the difference between the arc (length \(h\)) and \(\sin h\), that same arc straightened out vertically, fades away as the angle shrinks. So the ratio of the two lengths, \(\sin h / h\), draws ever closer to 1.
  • Formula: this is the limit that holds up all of trigonometric differentiation.
$$\lim_{h\to 0}\frac{\sin h}{h} = 1$$

Numbers make it real. At \(h=0.20\) this ratio is \(0.9933\), at \(h=0.10\) it is \(0.9983\), at \(h=0.05\) it is \(0.9996\) — each time the angle is halved, it steadily approaches 1.

The remaining one, \(B = \lim_{h\to 0}\dfrac{\cos h - 1}{h}\), is pulled out by this same engine. Multiplying top and bottom by \((\cos h + 1)\) and tidying gives \(\dfrac{\cos h - 1}{h} = -\dfrac{\sin h}{h}\cdot\dfrac{\sin h}{\cos h + 1}\); as \(h\to 0\), the front part is the 1 we just saw and the \(\sin h\) in the back goes to 0, so the product is entirely 0. Hence \(B=0\). With both parts filled in, the formula from Section 2 closes cleanly to \(\cos x\).

One bonus. Had we measured the angle in degrees rather than radians, the arc length would not be \(h\), so \(\sin h / h\) would converge not to 1 but to an awkward constant (\(\pi/180\)), and that messy factor would trail along in every derivative. The reason we measure trigonometric functions in radians lies right here — in keeping differentiation clean.

4. Check it yourself in the interactive below

Handle two scenes. One is the slope of sine tracing out cosine; the other is the engine that holds it up.

  • [Slope draws cos] mode: move the point left and right on the upper sine curve, and the tangent (slope) at that point tilts along with it. That slope value is plotted as a point in the lower panel, and the curve the point traces as it moves is exactly cosine. Where sine is steepest (\(x=0\)) the slope is 1; at sine's crest and trough (\(x=\pm\pi/2\)) the slope is 0 — confirm that it always matches the height of the cosine below.
  • [Why sin h/h → 1] mode: shrink the angle \(h\) with the slider, and the two bars — the unit circle's arc length \(h\) (amber) and the vertical shadow \(\sin h\) (green) — increasingly overlap into the same length. Watch the ratio \(\sin h / h\) approach 1 in the numbers. This is the one engine that closes the derivative of sine into cosine.
The slope of sin draws cos, and for a small angle arc and straight line become the same
In the first mode, moving the point on sine keeps its tangent slope matching the cosine height below. In the second mode, shrinking the angle makes the ratio of arc to sine converge to one.

Handling it, you feel with your fingertips that the derivative of sine being cosine is no coincidental resemblance but the relationship between two shadows of one rotation seen vertically and horizontally — and that the key to closing that relationship into a formula is one single fact: for a small angle, arc and straight line become the same.

5. Summary — two curves out of one rotation

  • Two shadows of a rotation: for a point going around the unit circle at steady speed, the height is \(\sin x\) and the horizontal position is \(\cos x\). Since the speed at which the height changes (the vertical component of the velocity arrow) is \(\cos x\), we have \(\dfrac{d}{dx}\sin x = \cos x\).
  • Confirmed by the definition: plug into the definition and expand with the addition formula, and the result hinges on two limits, \(\sin h/h \to 1\) and \((\cos h - 1)/h \to 0\). If the first is 1 and the second is 0, the formula closes to \(\cos x\).
  • The one engine: \(\sin h / h \to 1\) is the fact that 'for a very small angle, the arc (length \(h\)) and the vertical shadow \(\sin h\) are indistinguishable.' The second limit follows from it too. This engine runs because the angle is measured in radians.

Sine and cosine are not two curves that run separately; they are two shadows one rotation casts, vertically and horizontally. This interlock — where the rate of change of one becomes the position of the other — is what shows up, through the lens of differentiation, as \((\sin x)' = \cos x\).

Closing

Today we built the derivative of sine from a single rotation. But the other big part foreshadowed in the product rule post — the chain rule for differentiating the layered structure of a function inside a function — we have not yet built as a picture. When another function sits inside sine, as in \(\sin(2x)\) or \(\sin(x^2)\), how do we differentiate it? Taking today's \((\sin x)' = \cos x\) as a part, we will next unpack that layered structure from a single picture.