The Derivative of sin(x) Is cos(x)

The most beautiful derivative result in calculus may be this:

$\dfrac{d}{dx}\bigl[\sin x\bigr] = \cos x$

It says that the slope of the sine curve at any point is exactly the value of the cosine function there. The two curves are intertwined: where sine reaches its peak, cosine is zero (the slope is flat). Where sine crosses zero on the way up, cosine is at its maximum (the slope is steepest).

The Physics — Just Limit Magic

Starting from the limit definition of the derivative:

$\dfrac{d}{dx}\bigl[\sin x\bigr] = \lim_{h\to 0}\dfrac{\sin(x + h) - \sin x}{h}$

Use the angle-addition identity $\sin(x + h) = \sin x \cos h + \cos x \sin h$:

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

$= \lim_{h\to 0}\left[\sin x \cdot \dfrac{\cos h - 1}{h} + \cos x \cdot \dfrac{\sin h}{h}\right]$

The two fundamental trig limits are $\displaystyle\lim_{h\to 0}\dfrac{\sin h}{h} = 1$ and $\displaystyle\lim_{h\to 0}\dfrac{\cos h - 1}{h} = 0$. Plugging in:

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

That's the proof in seven lines.

Step-by-Step Solution

Given: $f(x) = \sin x$. Find: the slope at $x = a$ for several values of $a$, and verify it matches $\cos a$.

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Step 1 — Apply the formula.

We've shown $f'(x) = \cos x$. So at any point $a$, the slope of $\sin x$ is $\cos a$.

Step 2 — Tabulate at strategic points.

• At $x = 0$: $\sin 0 = 0$, slope $= \cos 0 = 1$ → tangent rises at 45° ✓
• At $x = \pi/2$: $\sin(\pi/2) = 1$ (peak), slope $= \cos(\pi/2) = 0$ → tangent is horizontal ✓
• At $x = \pi$: $\sin\pi = 0$, slope $= \cos\pi = -1$ → tangent descends at 45° ✓
• At $x = 3\pi/2$: $\sin(3\pi/2) = -1$ (trough), slope $= \cos(3\pi/2) = 0$ → tangent is horizontal ✓
• At $x = 2\pi$: back to 0, slope $= 1$, the cycle repeats ✓

The pattern is clear: the slope of sine is maximum ($+1$) at the zero-crossings going up, zero at the peaks and troughs, and minimum ($-1$) at the zero-crossings going down. That's exactly the shape of $\cos x$, shifted so that its peak aligns with sine's steepest ascent.

Step 3 — Verify with a specific intermediate point: $x = \pi/4$.

$\sin(\pi/4) = \dfrac{\sqrt{2}}{2} \approx 0.707$

$\cos(\pi/4) = \dfrac{\sqrt{2}}{2} \approx 0.707$

So the slope of $\sin x$ at $x = \pi/4$ is approximately $0.707$. The tangent line at that point has equation:

$y = 0.707\,(x - \pi/4) + 0.707$

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Answer: $\dfrac{d}{dx}[\sin x] = \cos x$. The slope of the sine curve at any point $x$ is exactly the value of $\cos x$ at that same point. The two functions are derivatives of each other (in a chain: $\sin' = \cos$, $\cos' = -\sin$, $-\sin' = -\cos$, $-\cos' = \sin$, back to start after four steps).

Try It

• Slide the point along the sine curve — watch the cyan tangent line rotate.
• The lower panel shows $\cos x$ in pink. The yellow dot tracks the current slope (read off the upper curve as the tangent's steepness, and read off the lower curve as the value of cos at that x).
• Notice that wherever $\sin x$ reaches a peak, $\cos x$ crosses zero — the slope is flat at the top of a wave.
• Wherever $\sin x$ crosses zero going up, $\cos x$ is at its maximum ($+1$) — that's the steepest ascent.