Differentiating Trigonometric Functions

The gradient of sin x, read off its graph

With x in radians, y = sin x climbs through the origin at gradient 1, because $\frac{\sin h}{h}\to 1$(sin h)/h → 1 as $h\to 0$h → 0. At $x=\frac{\pi }{2}$x = π/2 it reaches its peak, where the tangent is flat and the gradient is 0. At $x=\pi$x = π it falls through zero at gradient −1, at $\frac{3\pi }{2}$3π/2 it is flat again at its lowest point, and at $2\pi$2π it climbs at gradient 1 once more.

Those gradients, 1, 0, −1, 0, 1, are the values of cos x at the same five points. Where sine is steepest, cosine is at a peak or a trough; where sine peaks, cosine crosses zero.

The gold curve y = sin x and the dashed curve y = cos x, for x from −0.5 to 6.5 radians. The gold line y = x is the tangent to sin x at the origin, where cos 0 = 1. At $x=\frac{\pi }{2}$x = π/2, sin x peaks at 1 and cos x is 0.

The derivative of sin x, from the limit

The addition formula gives sin(x + h) = sin x cos h + cos x sin h. So the difference quotient is $\frac{\sin \left(x+h\right)-\sin x}{h}=\sin x\times \frac{\cos h-1}{h}+\cos x\times \frac{\sin h}{h}$(sin(x + h) − sin x)/h = sin x × (cos h − 1)/h + cos x × (sin h)/h.

Two limits finish it. With h in radians, $\frac{\sin h}{h}\to 1$(sin h)/h → 1. And $\frac{\cos h-1}{h}\to 0$(cos h − 1)/h → 0: since $\cos h-1=-2{\sin }^2\left(\frac{h}{2}\right)$cos h − 1 = −2 sin²(h/2), the quotient is $-\sin \left(\frac{h}{2}\right)\times \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$−sin(h/2) × (sin(h/2))/(h/2), which tends to 0 × 1 = 0. At h = 0.1 the two quotients are 0.99833 and −0.04996; at h = 0.01 they are 0.999983 and −0.0050.

So the difference quotient tends to sin x × 0 + cos x × 1, and the derivative of sin x is cos x. Check at x = 1: the chord from 1 to 1.001 has gradient $\frac{\sin 1.001-\sin 1}{0.001}=0.53988$(sin 1.001 − sin 1)/0.001 = 0.53988, and cos 1 = 0.54030.

The derivative of cos x

The same steps run on cos(x + h) = cos x cos h − sin x sin h. The difference quotient is $\cos x\times \frac{\cos h-1}{h}-\sin x\times \frac{\sin h}{h}$cos x × (cos h − 1)/h − sin x × (sin h)/h, which tends to cos x × 0 − sin x × 1. So the derivative of cos x is −sin x.

The minus sign shows on the graph. Between 0 and $\pi$π, cos x falls from 1 to −1, so its gradient is negative, and sin x is positive there. Check at x = 1: the chord from 1 to 1.001 has gradient −0.84174, and −sin 1 = −0.84147.

Four steps round a cycle

Differentiate sin x and the result is cos x. Differentiate that and the result is −sin x; a constant factor of −1 stays where it is. Then −sin x gives −cos x, and −cos x gives sin x again.

So differentiating four times in a row returns to the start: sin x, cos x, −sin x, −cos x, sin x. Differentiating sin x ten times goes round the cycle twice, which is eight steps, and then two more, to −sin x.

A point at angle $\theta$θ on a circle of radius 1 has height $\sin \theta$sin θ. Turning at 1 radian per second, it climbs at $\cos \theta$cos θ per second, the second bar. At $\theta =60^{\circ }$θ = 60° the height is 0.87 and the climb is 0.5; at $\theta =0$θ = 0 the height is 0 and the climb is at its fastest.

Radians only

Both proofs used $\frac{\sin h}{h}\to 1$(sin h)/h → 1, which holds only for h in radians. An angle of h degrees is $\frac{\pi h}{180}$πh/180 radians, so with h in degrees, $\frac{\sin \left(h^{\circ }\right)}{h}$sin(h°)/h tends to $\frac{\pi }{180}=0.01745$π/180 = 0.01745 instead.

Then every derivative carries that factor: the derivative of sin(x°) with respect to x is $\left(\frac{\pi }{180}\right)\cos \left(x^{\circ }\right)$(π/180) cos(x°). At x = 60 that is 0.01745 × 0.5 = 0.0087 per degree, not 0.5. In calculus, the angle in sin x and cos x is in radians unless it says otherwise.

Multiples, sums and a coefficient on the angle

Constant multiples and sums differentiate term by term. y = 3 sin x + 2 cos x has derivative 3 cos x − 2 sin x, and at x = 0 that is 3 × 1 − 2 × 0 = 3.

A coefficient on the angle needs the chain rule. For y = sin 5x, the inner function is 5x and the outer function is sin, so $\frac{dy}{dx}=\cos 5x\times 5=5\cos 5x$dy/dx = cos 5x × 5 = 5 cos 5x. At x = 0.2 that is 5 cos 1 = 2.7015.

For $y=\cos \left(x^2\right)$y = cos(x²), the inner function is $x^2$x² and the outer function is cos, so $\frac{dy}{dx}=-\sin \left(x^2\right)\times 2x=-2x\sin \left(x^2\right)$dy/dx = −sin(x²) × 2x = −2x sin(x²), which is −2 sin 1 = −1.6829 at x = 1.

The usual mistakes

Giving the derivative of cos x as sin x. Cosine falls from its peak at 0 while sin x is positive, so the derivative is −sin x.

Putting a minus sign on the derivative of sin x. Sine climbs through the origin, so its gradient there is +1 = cos 0.

Leaving out the inner factor. sin 5x differentiates to 5 cos 5x, not cos 5x.

Working in degrees. The results $\sin x\to \cos x$sin x → cos x and $\cos x\to -\sin x$cos x → −sin x hold only with x in radians.

A crank pin on a wheel

In the application below, the height of a pin on a turning wheel is 10 sin x. Its derivative, 10 cos x, gives how fast the pin rises, and where 10 cos x = 0 the pin is momentarily at rest.