Why Is the Derivative of Sine Cosine? A Geometric Proof Explained

The derivative of sine is one of the fundamental results in calculus:

$d(\sin\theta)/d\theta=\cos\theta$

We usually learn this result through algebraic manipulation and trigonometric identities. However, there is another way to understand it: through the geometry of a point moving around a unit circle.

This geometric approach raises several interesting questions:

What exactly does Δθ represent?
Is Δθ an angle or a length?
Why does the chord of a circle approach its tangent?
Is angle OPQ actually 90°?
Why are the two angles marked α equal?
Why is α equal to 90° − θ?
How does this geometry eventually produce cos θ?

Let’s answer these questions step by step.

1. Start with the unit circle

Consider a circle with centre O and radius 1.

Choose a point P on its circumference. The radius OP makes an angle θ with the positive horizontal axis.

Because the circle has radius 1, the coordinates of P are:

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

This gives us two important relationships:

The horizontal coordinate of P is cos θ.
The vertical coordinate of P is sin θ.

Therefore, the sine function represents the vertical position of the point on the unit circle.

Our objective is to determine how quickly this vertical position changes when the angle changes.

2. What does Δθ actually represent?

Suppose P moves a little further around the circumference to a nearby point Q.

The original angle is θ, and the new angle is:

$\theta+\Delta\theta$

The angular displacement is:

$\Delta\theta$

It is important to understand that Δθ is an angular change, not a straight-line distance.

The vertical coordinate changes from:

$\sin\theta$

to:

$\sin(\theta+\Delta\theta)$

Consequently, the vertical displacement is:

$\Delta y=\sin(\theta+\Delta\theta)-\sin\theta$

Notice that Δy depends on both θ and Δθ.

The same angular movement can produce different vertical displacements depending on the starting position of P.

For example, moving through a small angle near the rightmost point of the circle produces a different vertical displacement from moving through the same angle near the top.

This is why the derivative depends on θ.

3. Why does Δθ equal the arc length on a unit circle?

There is an important relationship between angular displacement and circular arc length:

$s=r\Delta\theta$

Here, s represents the arc length, r is the radius, and Δθ is measured in radians.

Since the radius of our circle is 1:

$s=1\times\Delta\theta$

Therefore:

$s=\Delta\theta$

This does not mean that an angle and a length are the same physical quantity. It means that their numerical values are equal in this particular unit-circle setting.

This distinction becomes important when we compare the angular displacement with the straight-line distance between P and Q.

4. The difference between an arc and a chord

There are two ways to measure the separation between P and Q.

Arc PQ: The curved distance along the circumference.

Chord PQ: The straight-line distance connecting P and Q.

For a finite angular displacement, the arc is longer than the chord.

However, as Δθ becomes smaller, the difference between the two becomes increasingly insignificant.

Let s represent the arc length PQ.

In the limit:

$\lim_{\Delta\theta\to0}PQ/s=1$

Since the arc length equals Δθ on a unit circle:

$\lim_{\Delta\theta\to0}PQ/\Delta\theta=1$

This is one of the essential steps in the geometric proof.

It allows us to use the chord as an approximation to the arc length when taking the limit.

5. Is angle OPQ exactly 90°?

This is an important question because the geometric proof relies on the relationship between a chord and a tangent.

Consider triangle OPQ.

Both OP and OQ are radii of the same circle:

$OP=OQ=1$

Therefore, triangle OPQ is isosceles.

The angle at the centre is Δθ. Because the triangle is isosceles, its two base angles are equal.

Using the angle-sum property of a triangle:

$\angle OPQ+\angle OQP+\Delta\theta=180^\circ$

Since the two base angles are equal:

$2\angle OPQ+\Delta\theta=180^\circ$

Therefore:

$\angle OPQ=90^\circ-\Delta\theta/2$

This gives us a precise answer.

For a finite, non-zero Δθ, angle OPQ is not exactly 90°.

However, as Δθ approaches zero:

$\lim_{\Delta\theta\to0}\angle OPQ=90^\circ$

Why?

Because the chord PQ approaches the tangent at P. A fundamental property of a circle states that the tangent is perpendicular to the radius drawn to the point of contact.

Thus, in the limiting position:

$OP\perp\text{tangent at P}$

This is why we say that angle OPQ approaches 90°, rather than claiming that it is already 90° for a finite angular displacement.

6. Understanding the two angles marked α

Now let’s examine the two angles marked α in the video.

This is perhaps the most interesting part of the geometric construction.

Although both angles are marked α, they are formed by different pairs of lines.

The lower α

The lower α is formed by:

The radius OP, viewed from P towards O.
The downward vertical direction.

The radius OP makes an angle θ with the horizontal direction.

Since the horizontal and vertical directions are perpendicular, the lower angle α is complementary to θ.

Therefore:

$\alpha=90^\circ-\theta$

Here, complementary angles simply means that their sum is 90°.

The upper α

The upper α is formed by:

The chord PQ.
The horizontal direction pointing left from P.

For a finite Δθ, the chord PQ is not exactly tangent to the circle.

However, as Δθ approaches zero, the chord approaches the tangent at P.

Now consider two pairs of perpendicular directions:

First pair:

Radius OP.
Tangent at P.

These are perpendicular because a tangent is perpendicular to the radius at the point of contact.

Second pair:

Horizontal direction.
Vertical direction.

These are also perpendicular.

Here is the geometric property that connects the two angles:

When both lines forming an angle are rotated through the same amount, the angle between them remains unchanged.

For example, if two lines form an angle of 40°, rotating both lines by 90° preserves that 40° angle.

Applying this property to our diagram:

The radius and vertical direction form the lower α.
The tangent and horizontal direction form the corresponding upper α.

Therefore, the two acute angles are equal.

Since the chord approaches the tangent as Δθ approaches zero, the upper α approaches the lower α.

Thus:

$\text{Upper }\alpha\longrightarrow\text{Lower }\alpha$

And because the lower α equals 90° − θ:

$\text{Upper }\alpha\longrightarrow90^\circ-\theta$

An important distinction: The angle marked α is not θ. It is the complementary angle to θ in this first-quadrant configuration.

Also, α should not be confused with angle OPQ. Angle OPQ approaches 90°, whereas the two angles marked α approach 90° − θ.

7. How does the small triangle produce cosine?

Now consider the small right triangle formed by:

The chord PQ.
The vertical displacement Δy.
The horizontal displacement between P and the vertical projection of Q.

Its hypotenuse is PQ, and its vertical side is Δy.

The angle between the chord and the horizontal direction approaches:

$90^\circ-\theta$

Using the definition of sine:

$\sin\alpha=\text{Opposite side}/\text{Hypotenuse}$

Therefore, as Δθ approaches zero:

$\sin\alpha\longrightarrow\Delta y/PQ$

Substituting:

$\alpha\longrightarrow90^\circ-\theta$

we obtain:

$\sin(90^\circ-\theta)\longrightarrow\Delta y/PQ$

Using the complementary-angle identity:

$\sin(90^\circ-\theta)=\cos\theta$

Therefore:

$\Delta y/PQ\longrightarrow\cos\theta$

This is the crucial geometric result.

The vertical displacement divided by the small chord approaches the cosine of the original angle.

8. Connect the chord to the angular displacement

We have already established that:

$\lim_{\Delta\theta\to0}PQ/\Delta\theta=1$

We also established that:

$\lim_{\Delta\theta\to0}\Delta y/PQ=\cos\theta$

Now multiply these two ratios:

$\Delta y/\Delta\theta=(\Delta y/PQ)\times(PQ/\Delta\theta)$

Taking the limit:

$\lim_{\Delta\theta\to0}\Delta y/\Delta\theta=\cos\theta\times1$

Therefore:

$\lim_{\Delta\theta\to0}\Delta y/\Delta\theta=\cos\theta$

We have now established the geometric relationship needed for the derivative.

9. Complete the proof using the definition of a derivative

Recall that the vertical displacement is:

$\Delta y=\sin(\theta+\Delta\theta)-\sin\theta$

Substitute this into the limiting expression:

$\lim_{\Delta\theta\to0}[\sin(\theta+\Delta\theta)-\sin\theta]/\Delta\theta=\cos\theta$

But the expression on the left is precisely the definition of the derivative of sine.

Consequently:

$d(\sin\theta)/d\theta=\cos\theta$

The geometric proof is complete.

10. What does the result mean geometrically?

There is a particularly interesting interpretation of this result.

A point on the unit circle has coordinates:

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

When the point moves around the circle:

Its horizontal coordinate is cos θ.
Its vertical coordinate is sin θ.
Its instantaneous vertical displacement per unit of angular displacement is cos θ.

In other words:

$\text{Rate of change of vertical position}=\text{Horizontal coordinate}$

This is the geometric meaning of the derivative of sine.

The result is not merely an algebraic identity. It describes how a point moving around a circle changes its vertical position.

Common questions and doubts

Question	Explanation
Is Δθ a length?	No. It is an angular displacement measured in radians.
Why does Δθ equal the arc length?	Because the radius is 1 and arc length equals radius multiplied by angular displacement.
Is the chord equal to the arc?	Not for a finite angle. They become approximately equal as Δθ approaches zero.
Is angle OPQ exactly 90°?	No. It approaches 90° as Δθ approaches zero.
Why is the tangent perpendicular to OP?	A tangent is perpendicular to the radius at the point of contact.
Why are the two α angles equal?	In the limiting position, they are formed by two pairs of perpendicular directions.
Is α equal to θ?	No. In the first-quadrant diagram, α is 90° − θ.
Is angle OPQ the same as α?	No. Angle OPQ approaches 90°, while the two α angles approach 90° − θ.
Why must angles be in radians?	The arc-length relationship and the derivative formula require radians.

Final takeaway

The most important lesson from this proof is that we must understand what each algebraic symbol represents geometrically.

The two changes are different:

Δθ represents a small angular rotation.
Δy represents the resulting change in vertical position.

The unit circle connects these changes through its arc, chord, tangent and perpendicular geometry.

As the angular displacement approaches zero:

The chord approaches the tangent.
The chord and arc become almost equal.
Angle OPQ approaches 90°.
The two angles marked α approach 90° − θ.
The ratio of vertical displacement to angular displacement approaches cos θ.

The horizontal coordinate of the point therefore determines the instantaneous rate of change of its vertical coordinate.

Hence:

$d(\sin\theta)/d\theta=\cos\theta$

The derivative of sine is cosine because, at every point on the unit circle, the horizontal coordinate determines the instantaneous rate at which the vertical coordinate changes with respect to the angle.

Why Is the Derivative of Sine Cosine? A Geometric Proof Explained

1. Start with the unit circle

2. What does Δθ actually represent?

3. Why does Δθ equal the arc length on a unit circle?

4. The difference between an arc and a chord

5. Is angle OPQ exactly 90°?

6. Understanding the two angles marked α

The lower α

The upper α

7. How does the small triangle produce cosine?

8. Connect the chord to the angular displacement

9. Complete the proof using the definition of a derivative

10. What does the result mean geometrically?

Common questions and doubts

Final takeaway

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

1. Start with the unit circle

2. What does Δθ actually represent?

3. Why does Δθ equal the arc length on a unit circle?

4. The difference between an arc and a chord

5. Is angle OPQ exactly 90°?

6. Understanding the two angles marked α

The lower α

The upper α

7. How does the small triangle produce cosine?

8. Connect the chord to the angular displacement

9. Complete the proof using the definition of a derivative

10. What does the result mean geometrically?

Common questions and doubts

Final takeaway

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series