Differentiating sin(x) and cos(x)
Key Terms
- Derivative of sin x
- d/dx[sin x] = cos x (x in radians).
- Derivative of cos x
- d/dx[cos x] = −sin x (x in radians).
- Derivative of tan x
- d/dx[tan x] = sec² x.
- Chain rule for trig
- d/dx[sin(f(x))] = f′(x) cos(f(x)); d/dx[cos(f(x))] = −f′(x) sin(f(x)).
- Radians essential
- Derivative rules for trig functions are ONLY valid when x is in radians; convert degrees first.
- Higher derivatives
- d²/dx²[sin x] = −sin x; the pattern of derivatives repeats with period 4.
d/dx[sin x] = cos x
d/dx[cos x] = −sin x
d/dx[tan x] = sec²x = 1/cos²x
Higher Derivatives — the cycle:
sin x → cos x → −sin x → −cos x → sin x → …
d²/dx²[sin x] = −sin x
d4/dx4[sin x] = sin x (period 4)
Finding zeros of gradient:
f(x) = sin x: f′(x) = cos x = 0 when x = π/2 + nπ
f(x) = cos x: f′(x) = −sin x = 0 when x = nπ
dy/dx = 3 cos x − 2(−sin x) = 3 cos x + 2 sin x
dy/dx = cos x = 0
cos x = 0 when x = π/2 and x = 3π/2 on [0, 2π].
These correspond to the maximum and minimum of sin x.
Why Radians? The Geometric Derivation
The derivative of sin x = cos x is one of the most beautiful results in calculus — but it only holds when x is in radians. To understand why, consider the unit circle. As the angle x (in radians) increases by a tiny amount δx, the point (cos x, sin x) moves along the circle. The rate of change of the vertical coordinate sin x with respect to x is precisely the cosine of the angle — that is, cos x. This geometric argument gives d/dx[sin x] = cos x.
The formal limit definition confirms this: limδx→0 [sin(x+δx) − sin x]/δx = cos x, using the key limits limθ→0(sin θ)/θ = 1 and limθ→0(cos θ − 1)/θ = 0. Both of these limits only equal these values when θ is in radians.
The Cycle of Trig Derivatives
One of the most elegant features of sin x and cos x is that their derivatives cycle with period 4:
sin x → cos x → −sin x → −cos x → sin x → …
This means the fourth derivative of sin x is sin x again. To find the nth derivative, divide n by 4 and check the remainder: remainder 0 → sin x; remainder 1 → cos x; remainder 2 → −sin x; remainder 3 → −cos x.
The second derivative d²y/dx²[sin x] = −sin x is especially important: it tells us that y = sin x satisfies the differential equation y′′ = −y. This is the equation of simple harmonic motion — every oscillating system, from pendulums to springs to sound waves, is governed by this equation.
Stationary Points and Gradient Analysis
For y = sin x, the gradient is dy/dx = cos x. The gradient is zero where cos x = 0, i.e. at x = π/2 + nπ for integer n. At x = π/2, the sin function reaches its maximum (y = 1); at x = 3π/2, it reaches its minimum (y = −1).
For mixed functions like y = 2 sin x + x, the gradient is dy/dx = 2 cos x + 1. Setting this to zero gives cos x = −1/2, so x = 2π/3 or x = 4π/3 on [0, 2π]. The interaction of the trig and linear terms produces stationary points that are not the obvious maxima or minima of sin x alone.
Physical Interpretation: Velocity of a Wave
If a particle’s displacement is given by y = A sin(t), then its velocity is dy/dt = A cos(t). The velocity is maximum when cos t = 1 (at t = 0), where the displacement is zero. The velocity is zero when cos t = 0 (at t = π/2), where the displacement is at its maximum. This captures the physics of oscillation: at the equilibrium position, the particle moves fastest; at the extreme positions, it momentarily stops.
Mastery Practice
- Differentiate y = 4 sin x + 3 cos x.
- Differentiate y = 2 cos x − 5 sin x.
- Find d²y/dx² for y = 3 sin x.
- Find d10/dx10[cos x].
- Find the gradient of y = sin x + cos x at x = π/4.
- Find all x ∈ [0, 2π] where f′(x) = 0 for f(x) = 2 cos x.
- Find the stationary points of y = 2 sin x + x on [0, 2π] and determine their nature.
- Find the equation of the tangent to y = sin x at x = π/6.
- Show that y = sin x satisfies the equation d²y/dx² + y = 0.
- A particle’s displacement is x(t) = 3 sin t − cos t metres. Find the velocity and acceleration functions. Find when the velocity is zero for t ∈ [0, 2π], and the displacement at those times.