Navigation: Wiki Home > Skills > Derivatives of Sine and Cosine
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.
| Property | Value |
|---|---|
| Concept | Trigonometric Derivatives |
| Chapter | 2.4 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
$$\boxed{\frac{d}{dx}(\sin x) = \cos x}$$
$$\boxed{\frac{d}{dx}(\cos x) = -\sin x}$$
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 $\pi/180$.
Before we can prove these formulas, we need two special limits:
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \qquad \lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0$$
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.
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 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.
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.
Find $\frac{d}{dx}(5\sin x - 3\cos x)$.
Find $\frac{d}{dx}(x^3 \sin x)$.
Find the equation of the tangent line to $y = \sin x + \cos x$ at the point where $x = 0$.
Find $\frac{d^{83}}{dx^{83}}(\cos x)$.
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.
The graph below shows $f(x) = \sin x$.
1 | ∩
| / \
0 |--/-----\-----/--
| \ /
-1 | ∪
0 π/2 π 3π/2 2π
At which point(s) is $f'(x) = 0$?
If $f(x) = \cos x$, then on the interval $(0, \pi)$:
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.
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