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Pure · Differentiation

Chapter 1 · 4

The idea

Differentiating trigonometric functions

The six standard trig derivatives — sin→cos, cos→−sin, tan→sec², sec→sec tan, cosec→−cosec cot, cot→−cosec² — why x must be in radians, and combining them with the chain, product and quotient rules.

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Pure · Differentiation

Differentiating trigonometric functions

The six standard trig derivatives — sin→cos, cos→−sin, tan→sec², sec→sec tan, cosec→−cosec cot, cot→−cosec² — why x must be in radians, and combining them with the chain, product and quotient rules.

Why it works

Sine, from first principles

The derivative of sin⁡x\sin x is the limit of sin⁡(x+h)−sin⁡xh\dfrac{\sin(x+h) - \sin x}{h} as h→0h \to 0. Expand the top with the addition formula sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h) = \sin x\cos h + \cos x\sin h: sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xh=sin⁡x⋅cos⁡h−1h+cos⁡x⋅sin⁡hh.\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}.

The small-angle finish

Now use the small-angle results for tiny hh (in radians): sin⁡h≈h\sin h \approx h, so sin⁡hh→1\frac{\sin h}{h}\to 1; and cos⁡h≈1−12h2\cos h \approx 1 - \tfrac12 h^2, so cos⁡h−1h≈−12h→0\frac{\cos h - 1}{h}\approx -\tfrac12 h \to 0. The first term vanishes and the second leaves cos⁡x\cos x: ddxsin⁡x=cos⁡x.\frac{d}{dx}\sin x = \cos x. The same calculation on cos⁡x\cos x gives ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x = -\sin x (note the minus — cosine is falling where sine is positive).

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