Derivatives of Sine and Cosine

We have (sin x)′ = cos x and (cos x)′ = −sin x, each for x in radians. Together with the second derivatives this forms a cycle of four: sin → cos → −sin → −cos and back to sin. After four differentiations you are back at the start. The minus appears exactly at the transition from cosine: the cosine starts at its maximum at x = 0, then falls, so its slope must start negative, and −sin x does exactly that. Conversely the sine starts at 0 with maximal slope 1, and cos 0 = 1 fits. Once you trace the cycle on the graphs, you mix up the signs far less often.

The rule (sin x)′ = cos x rests on the limit sin(h)/h → 1 as h → 0, and this holds only in radians, where an angle is measured directly as arc length on the unit circle. If you work in degrees, the limit is π/180 ≈ 0.01745 instead, and every derivative acquires this factor: (sin x°)′ = (π/180)·cos x°. The elegant relations like (sin x)″ = −sin x would be lost. Practical consequence: always work in radians in calculus and set the calculator to RAD. A typical exam error is an extremum at "x = 90" instead of x = π/2 because the calculator was in DEG mode.

Use the chain rule: the outer function sin becomes cos, the inner function bx + c has derivative b. So (a·sin(bx + c))′ = a·b·cos(bx + c). The amplitude a is kept as a factor, the angular frequency b enters multiplicatively, the phase shift c stays unchanged in the argument. Example: f(x) = 3·sin(2x − π) has f′(x) = 6·cos(2x − π). Physically the factor b means: a faster oscillation with the same amplitude has a proportionally larger maximum velocity. The standard mistake is forgetting b; when in doubt, check with a point, for example the initial slope at x = 0.

Via the derivatives: extrema of sin x lie where cos x = 0, that is at x = π/2 + k·π; inflection points of sin x lie where the second derivative −sin x vanishes, that is exactly at the zeros x = k·π of the sine itself. So the curve inflects at every axis crossing and is steepest there with slope ±1. For the cosine everything is shifted by π/2: maxima and minima at x = k·π, inflection points at x = π/2 + k·π. This regularity makes trigonometric curve analysis very predictable: once you have one feature, you know all others by shifting quarter periods.

The tangent is the quotient tan x = sin x/cos x. The quotient rule gives (tan x)′ = (cos x·cos x − sin x·(−sin x))/cos²x = (cos²x + sin²x)/cos²x. With the trigonometric Pythagoras sin²x + cos²x = 1 this simplifies to (tan x)′ = 1/cos²x, equivalently 1 + tan²x. Since 1/cos²x is always positive, the tangent increases strictly on each of its branches; at the poles x = π/2 + k·π, where cos x = 0, it does not exist. The form 1 + tan²x is useful in physics and integration, for instance when derivatives are to be expressed through tan itself.