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Derivatives of Sine and Cosine

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Reference: Stewart §2.4

Why These Two Derivatives Matter

Every oscillating system in nature (a swinging pendulum, a vibrating guitar string, an alternating current) is described by sine and cosine functions. To understand how fast these systems change at any moment, we need their derivatives.

Here’s the beautiful surprise: the derivative of sine is cosine, and the derivative of cosine is negative sine. These two functions are intertwined in a cycle that repeats every four derivatives. This elegant pattern makes trigonometric derivatives some of the most memorable in all of calculus.

Prerequisite Map

Quick Reference

Property Value
Concept Trigonometric Derivatives
Chapter 2.4
Difficulty Intermediate
Time ~20 minutes

Key Formulas

$$\boxed{\frac{d}{dx}(\sin x) = \cos x}$$

$$\boxed{\frac{d}{dx}(\cos x) = -\sin x}$$

Critical reminder: These formulas are only valid when $x$ is measured in radians. If you work in degrees, you’ll get an extra factor of $\pi/180$.

The Two Essential Limits

Before we can prove these formulas, we need two special limits:

$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \qquad \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0$$

Why $\frac{\sin \theta}{\theta} \to 1$?

For small angles (in radians), the sine of the angle is approximately equal to the angle itself:

        Arc length = θ (on unit circle)
              ↗
             /
            /|
           / |  Height = sin θ
          /  |
         /θ__|
        O

For small θ: sin θ ≈ θ

The Squeeze Theorem makes this rigorous: $\cos \theta < \frac{\sin \theta}{\theta} < 1$ for small positive $\theta$, and both bounds approach 1.

Proof: Derivative of Sine

Starting from the definition:

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

Using the angle addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$:

$$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$

$$= \lim_{h \to 0} \left( \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h} \right)$$

$$= \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$

The Cyclic Pattern

The derivatives of sine and cosine repeat in a cycle of four:

$n$ $\frac{d^n}{dx^n}(\sin x)$ $\frac{d^n}{dx^n}(\cos x)$
0 $\sin x$ $\cos x$
1 $\cos x$ $-\sin x$
2 $-\sin x$ $-\cos x$
3 $-\cos x$ $\sin x$
4 $\sin x$ $\cos x$

Pattern: Differentiating four times returns you to the original function.

Finding the $n$th derivative: Divide $n$ by 4 and look at the remainder.

Physical Interpretation

For simple harmonic motion with position $s(t) = A\sin(t)$:

Notice: $a = -s$. The acceleration is proportional to position but opposite in sign. This is the defining property of simple harmonic motion.

Position:      ∿∿∿∿∿  (sine wave)
                 ↓ derivative
Velocity:      ∿∿∿∿∿  (cosine = shifted sine)
                 ↓ derivative
Acceleration:  ∿∿∿∿∿  (negative sine)

When position is at maximum, velocity is zero. When position crosses zero, velocity is at maximum.

Practice Problems

Level 1 Direct Differentiation

Find $\frac{d}{dx}(5\sin x - 3\cos x)$.

Thought Process

This is a linear combination of sine and cosine. Apply the derivative to each term separately, remembering that constants pull out:

  • Derivative of $\sin x$ is $\cos x$
  • Derivative of $\cos x$ is $-\sin x$ (note the negative!)
Show Answer

$$\frac{d}{dx}(5\sin x - 3\cos x) = 5\cos x - 3(-\sin x) = 5\cos x + 3\sin x$$

Level 2 Product Rule with Sine

Find $\frac{d}{dx}(x^3 \sin x)$.

Thought Process

This is a product of two functions: $f(x) = x^3$ and $g(x) = \sin x$.

Apply the product rule: $(fg)' = f'g + fg'$

We need:

  • $f'(x) = 3x^2$
  • $g'(x) = \cos x$
Show Answer

Using the product rule: $$\frac{d}{dx}(x^3 \sin x) = (3x^2)(\sin x) + (x^3)(\cos x)$$ $$= 3x^2 \sin x + x^3 \cos x$$

This can be factored as $x^2(3\sin x + x\cos x)$.

Level 3 Tangent Line to a Trig Curve

Find the equation of the tangent line to $y = \sin x + \cos x$ at the point where $x = 0$.

Thought Process

For a tangent line, we need:

  1. The point: evaluate $y$ at $x = 0$
  2. The slope: evaluate $y'$ at $x = 0$

Then use point-slope form: $y - y_1 = m(x - x_1)$

Remember: $\sin 0 = 0$ and $\cos 0 = 1$

Show Answer

Step 1: Find the point. At $x = 0$: $y = \sin 0 + \cos 0 = 0 + 1 = 1$ Point: $(0, 1)$

Step 2: Find the slope. $y' = \cos x - \sin x$ At $x = 0$: $y'(0) = \cos 0 - \sin 0 = 1 - 0 = 1$

Step 3: Write the equation. $y - 1 = 1(x - 0)$ $$y = x + 1$$

Level 4 Finding the 83rd Derivative

Find $\frac{d^{83}}{dx^{83}}(\cos x)$.

Thought Process

The derivatives of cosine cycle with period 4:

  • $\cos x \to -\sin x \to -\cos x \to \sin x \to \cos x \to \ldots$

To find the 83rd derivative, divide 83 by 4 and look at the remainder:

  • $83 = 4 \times 20 + 3$
  • Remainder is 3, so we want the 3rd derivative of $\cos x$
Show Answer

The cycle: $\cos x \xrightarrow{1} -\sin x \xrightarrow{2} -\cos x \xrightarrow{3} \sin x \xrightarrow{4} \cos x$

Since $83 = 4(20) + 3$, the remainder is 3.

The 3rd derivative of $\cos x$ is $\sin x$.

$$\frac{d^{83}}{dx^{83}}(\cos x) = \sin x$$

Level 5 Proving the Cosine Derivative Formula

Prove that $\frac{d}{dx}(\cos x) = -\sin x$ using the definition of the derivative.

Hint: Use the angle addition formula $\cos(x+h) = \cos x \cos h - \sin x \sin h$ and the two special limits.

Thought Process

Follow the same structure as the sine proof:

  1. Write the limit definition
  2. Apply the angle addition formula for cosine
  3. Rearrange to isolate the two special limits
  4. Substitute the limit values

The key difference: cosine addition has a minus sign, which is where the negative in $-\sin x$ comes from.

Show Answer

Step 1: Definition of derivative. $$\frac{d}{dx}(\cos x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$$

Step 2: Apply angle addition formula. $$= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$$

Step 3: Regroup terms. $$= \lim_{h \to 0} \frac{\cos x(\cos h - 1) - \sin x \sin h}{h}$$

$$= \lim_{h \to 0} \left( \cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h} \right)$$

Step 4: Apply the special limits. $$= \cos x \cdot 0 - \sin x \cdot 1$$

$$= -\sin x$$

Therefore: $\frac{d}{dx}(\cos x) = -\sin x$ $\square$

CCI-Style Conceptual Questions

CCI Graph Interpretation

The graph below shows $f(x) = \sin x$.

     1 |     ∩
       |   /   \
     0 |--/-----\-----/--
       |         \   /
    -1 |          ∪
       0   π/2   π  3π/2  2π

At which point(s) is $f'(x) = 0$?

  1. At $x = 0$ and $x = \pi$
  2. At $x = \pi/2$ and $x = 3\pi/2$
  3. At $x = \pi$ only
  4. At $x = 0, \pi/2, \pi, 3\pi/2$
Thought Process

$f'(x) = 0$ where the tangent line is horizontal, at the peaks and valleys of the sine curve.

Looking at the graph:

  • Peak at $x = \pi/2$ (horizontal tangent)
  • Valley at $x = 3\pi/2$ (horizontal tangent)

At $x = 0$ and $x = \pi$, the curve is crossing zero with maximum slope (steepest), so $f'(x) \neq 0$ there.

Show Answer

Answer: (B) At $x = \pi/2$ and $x = 3\pi/2$

$f'(x) = \cos x = 0$ when $x = \pi/2, 3\pi/2, 5\pi/2, \ldots$

These are exactly the points where the sine curve has horizontal tangents (maxima and minima).

CCI Sign of the Derivative

If $f(x) = \cos x$, then on the interval $(0, \pi)$:

  1. $f'(x) > 0$ for all $x$ in $(0, \pi)$
  2. $f'(x) < 0$ for all $x$ in $(0, \pi)$
  3. $f'(x) > 0$ for $x$ in $(0, \pi/2)$ and $f'(x) < 0$ for $x$ in $(\pi/2, \pi)$
  4. $f'(x) = 0$ for all $x$ in $(0, \pi)$
Thought Process

Think about what $\cos x$ looks like on $(0, \pi)$:

  • At $x = 0$: $\cos 0 = 1$ (maximum)
  • At $x = \pi$: $\cos \pi = -1$ (minimum)

The function is decreasing throughout this interval, going from 1 down to -1.

A decreasing function has a negative derivative.

Alternatively: $f'(x) = -\sin x$, and $\sin x > 0$ for all $x$ in $(0, \pi)$, so $f'(x) = -\sin x < 0$.

Show Answer

Answer: (B) $f'(x) < 0$ for all $x$ in $(0, \pi)$

Since $f'(x) = -\sin x$ and $\sin x > 0$ on $(0, \pi)$, we have $f'(x) < 0$ throughout this interval.

This matches the fact that $\cos x$ is strictly decreasing from 1 to -1 on $(0, \pi)$.


Common Misconceptions

Common misconception

the derivative formulas $(\sin x)' = \cos x$ and $(\cos x)' = -\sin x$ work regardless of the angle unit used.

This is the radian-as-arc-length-in-radius-units error. Radians are defined as a dimensionless ratio (arc length divided by radius), and the limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ holds only when $\theta$ is measured in radians. When $x$ is in degrees, the same limit evaluates to $\pi/180$, and the chain rule introduces that factor: $\frac{d}{dx}[\sin(x^\circ)] = \frac{\pi}{180}\cos(x^\circ)$. Using $(\sin x)' = \cos x$ in degree measure gives an answer off by a factor of approximately $0.01745$.

Common misconception

$\sin$ and $\cos$ are symbols that can be separated from their argument like multipliers.

This is the trig-as-algebra-symbols error. Students sometimes write $\sin(x + h) = \sin x + \sin h$, treating $\sin$ as a factor that distributes over addition. The correct expansion requires the angle addition formula: $\sin(x + h) = \sin x \cos h + \cos x \sin h$. The distinction matters in the proof of $(\sin x)' = \cos x$: the derivation uses the addition formula at the key step, and treating sine as distributive makes the proof collapse entirely.


Mastery Checklist

Mental Model

The Phase Shift Connection:

Think of $\cos x$ as a sine wave shifted left by $\pi/2$:

$$\cos x = \sin\left(x + \frac{\pi}{2}\right)$$

When you differentiate sine, you get cosine, a function that is $\pi/2$ ahead in the wave cycle. Differentiating again shifts another $\pi/2$, giving $-\sin x$ (which is $\pi$ ahead, i.e., the negative).

Four shifts of $\pi/2$ brings you full circle: $4 \times \frac{\pi}{2} = 2\pi$, which is one complete period.


Connections

Looking back:

Looking ahead:

Real-world connections:



Last updated: 2026-01-22