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Why the derivative of sin(x) is cos(x)

Visualize sine as the height of a point on the unit circle; a tiny nudge in angle changes that height at a rate equal to the horizontal position — cos(x).

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Sine as height
y=sin⁡(θ)y = \sin(\theta)
Picture a point moving around the unit circle. Its vertical height at angle theta is sin(theta).
Step-by-step solver
1

Start from the definition

The derivative measures the rate of change of height as the angle changes.

ddθsin⁡(θ)=lim⁡h→0sin⁡(θ+h)−sin⁡(θ)h\frac{d}{d\theta}\sin(\theta) = \lim_{h\to0}\frac{\sin(\theta+h)-\sin(\theta)}{h}
2

Apply the angle sum identity

Expand using sin(theta+h) = sin(theta)cos(h) + cos(theta)sin(h).

sin⁡(θ+h)=sin⁡θcos⁡h+cos⁡θsin⁡h\sin(\theta+h)=\sin\theta\cos h+\cos\theta\sin h
3

Use small-angle limits

As h approaches zero, cos(h) approaches 1 and sin(h)/h approaches 1.

lim⁡h→0sin⁡hh=1,lim⁡h→0cos⁡h=1\lim_{h\to0}\frac{\sin h}{h}=1,\quad \lim_{h\to0}\cos h=1
4

Simplify

The sin(theta) terms cancel, leaving cos(theta).

ddθsin⁡(θ)=cos⁡(θ)\frac{d}{d\theta}\sin(\theta)=\cos(\theta)

Original question

Explain why the derivative of sin(x) is cos(x), intuitively.

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