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Derivatives of Sine and Cosine

MATH161
Reference: Stewart 2.4  •  Chapter: 2  •  Section: 4

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Derivatives of Sine and Cosine

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

This skill
Derivatives of Sine & Cosine

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 $(0, \pi)$.

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's $\pi/2$ "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:


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Basic Differentiation Rules Skills Index Derivatives of Other Trig Functions

Last updated: 2026-01-22