Differentiation from First Principles (sin x and cos x) — Edexcel A-level Mathematics (Pure)

SPEC 7.1 • DIFFERENTIATION

FIRST PRINCIPLES: sin x AND cos x

Proving the derivatives of trigonometric functions

Why Radians Matter

Critical

The derivatives of sin x and cos x ONLY work when x is measured in radians.

When we differentiate trigonometric functions, we rely on certain limit results that are only true when angles are in radians.

Key Point

If you use degrees, you will get different (incorrect) results. Always check that your calculator is in radian mode for calculus questions.

In this lesson, we'll prove from first principles that:

\frac{d}{d x} (sin x) = cos x

\frac{d}{d x} (cos x) = - sin x

Key Limit Results

To prove the derivatives of sin x and cos x, we need these essential limit results:

Key Limits

h \to 0 lim \frac{sin h}{h} = 1

h \to 0 lim \frac{cos h - 1}{h} = 0

Given in Formula Book

These limits are provided in the formula booklet. You need to know how to use them, not prove them.

These results come from the small angle approximations: as h → 0, sin h ≈ h and cos h ≈ 1.

Compound Angle Formulae

We'll need these formulae to expand sin(x + h) and cos(x + h):

Formulae

sin (A + B) = sin A cos B + cos A sin B

cos (A + B) = cos A cos B - sin A sin B

Application

For our proofs:

sin (x + h) = sin x cos h + cos x sin h

cos (x + h) = cos x cos h - sin x sin h

Worked Example01

Prove from first principles that if $f (x) = sin x$ , then $f^{'} (x) = cos x$ .

f^{'} (x) = h \to 0 lim \frac{sin ( x + h ) - sin x}{h}

Using sin(A + B) = sin A cos B + cos A sin B:

= h \to 0 lim \frac{sin x cos h + cos x sin h - sin x}{h}

Rearranging by factoring out sin x:

= h \to 0 lim \frac{sin x ( cos h - 1 ) + cos x sin h}{h}

= h \to 0 lim [sin x \cdot \frac{cos h - 1}{h} + cos x \cdot \frac{sin h}{h}]

Using the standard limits:

= sin x \cdot 0 + cos x \cdot 1 = cos x

Therefore, $\frac{d}{d x} (sin x) = cos x$

Worked Example02

Prove from first principles that if $f (x) = cos x$ , then $f^{'} (x) = - sin x$ .

f^{'} (x) = h \to 0 lim \frac{cos ( x + h ) - cos x}{h}

Using cos(A + B) = cos A cos B - sin A sin B:

= h \to 0 lim \frac{cos x cos h - sin x sin h - cos x}{h}

Rearranging (note the minus sign):

= h \to 0 lim \frac{cos x ( cos h - 1 ) - sin x sin h}{h}

= h \to 0 lim [cos x \cdot \frac{cos h - 1}{h} - sin x \cdot \frac{sin h}{h}]

Using the standard limits:

= cos x \cdot 0 - sin x \cdot 1 = - sin x

Therefore, $\frac{d}{d x} (cos x) = - sin x$ (Don't forget the negative!)

Summary of Results

Derivative of sin x

\frac{d}{d x} (sin x) = cos x

Derivative of cos x

\frac{d}{d x} (cos x) = - sin x

Memory Aid

"Sine goes to Cosine, Cosine goes to Negative Sine"

Or think of it as a cycle: sin → cos → -sin → -cos → sin ...

Common Error

The most common mistake is forgetting the negative sign when differentiating cos x.

03Worked Example

The curve C has equation $y = sin x + cos x$ for $0 \leq x \leq 2 π$ .

(a) Find $\frac{d y}{d x}$

(b) Find the x-coordinates of the stationary points of C.

(a)

y = sin x + cos x

\frac{d y}{d x} = cos x - sin x

(b) At stationary points,

\frac{d y}{d x} = 0

:

cos x - sin x = 0

cos x = sin x

tan x = 1

In the interval

[0, 2 π]

:

x = \frac{π}{4} or x = \frac{5 π}{4}

Worked Example04

Find the equation of the tangent to the curve $y = cos x$ at the point where $x = \frac{π}{6}$ .

Step 1: Find the y-coordinate

When

x = \frac{π}{6}

:

y = cos \frac{π}{6} = \frac{3}{2}

Step 2: Find the gradient

\frac{d y}{d x} = - sin x

At

x = \frac{π}{6}

: gradient

= - sin \frac{π}{6} = - \frac{1}{2}

Step 3: Form the equation using

y - y_{1} = m (x - x_{1})

y - \frac{3}{2} = - \frac{1}{2} (x - \frac{π}{6})

y = - \frac{1}{2} x + \frac{π}{12} + \frac{3}{2}

Practice

Question 1: Write out the first principles proof that

\frac{d}{d x} (cos x) = - sin x

, stating clearly the two limit results you use.

f^{'} (x) = h \to 0 lim \frac{cos ( x + h ) - cos x}{h}

Using cos(A + B) = cos A cos B - sin A sin B:

= h \to 0 lim \frac{cos x cos h - sin x sin h - cos x}{h}

= h \to 0 lim [cos x \cdot \frac{cos h - 1}{h} - sin x \cdot \frac{sin h}{h}]

Using:

h \to 0 lim \frac{sin h}{h} = 1

and

h \to 0 lim \frac{cos h - 1}{h} = 0

= cos x \cdot 0 - sin x \cdot 1 = - sin x

Practice

Question 2: The curve C has equation

y = 2 sin x - cos x

.

(a) Find

\frac{d y}{d x}

(b) Find the gradient of C at the point where

x = \frac{π}{3}

(a)

y = 2 sin x - cos x

\frac{d y}{d x} = 2 cos x - (- sin x) = 2 cos x + sin x

(b) At

x = \frac{π}{3}

:

Gradient = 2 cos \frac{π}{3} + sin \frac{π}{3} = 2 \times \frac{1}{2} + \frac{3}{2} = 1 + \frac{3}{2}

Practice

Question 3: Find the equation of the normal to the curve

y = sin x

at the point where

x = \frac{π}{4}

.

Step 1: Find the point

When

x = \frac{π}{4}

:

y = sin \frac{π}{4} = \frac{2}{2}

Step 2: Find gradient of tangent

\frac{d y}{d x} = cos x

, so at

x = \frac{π}{4}

: gradient of tangent

= cos \frac{π}{4} = \frac{2}{2}

Step 3: Gradient of normal =

- \frac{1}{gradient of tangent} = - \frac{2}{2} = - 2

Step 4: Equation:

y - \frac{2}{2} = - 2 (x - \frac{π}{4})

y = - 2 x + \frac{π 2}{4} + \frac{2}{2}

Summary

Key Results

Function f(x)Derivative f'(x)

sin x

cos x

cos x

- sin x

Key Limits (Given in Formula Book)

h \to 0 lim \frac{sin h}{h} = 1

h \to 0 lim \frac{cos h - 1}{h} = 0

Exam Tips

These proofs are commonly examined - learn the structure!

Always use radians for trigonometric differentiation

Don't forget the negative sign in d/dx(cos x) = -sin x

State the limit results clearly when proving from first principles