Trigonometric Proofs — Edexcel A-level Mathematics (Pure)

SPEC 5.6, 5.7 • TRIGONOMETRY

Constructing Trigonometric Proofs

Master the art of proving trigonometric identities using algebraic techniques and key identities

What is a Trigonometric Proof?

A trigonometric proof shows that two expressions are identical for all valid values of the variable.

Identity vs Equation

Identity: True for ALL values (use ≡ symbol)

sin^{2} θ + cos^{2} θ \equiv 1

is true for every θ

Equation: True for SOME values (use = symbol)

sin θ = 0.5

is only true when θ = 30°, 150°, etc.

Golden Rule

In a proof, work on ONE SIDE ONLY and transform it to match the other side exactly.

Key Identities Reference

Pythagorean Identities:

sin^{2} θ + cos^{2} θ \equiv 1

1 + tan^{2} θ \equiv sec^{2} θ

1 + cot^{2} θ \equiv csc^{2} θ

Quotient Identities:

tan θ \equiv \frac{s i n θ}{c o s θ}

cot θ \equiv \frac{c o s θ}{s i n θ}

Reciprocal Identities:

sec θ \equiv \frac{1}{c o s θ}

csc θ \equiv \frac{1}{s i n θ}

cot θ \equiv \frac{1}{t a n θ}

Rearrangements:

1 - sin^{2} θ \equiv cos^{2} θ

1 - cos^{2} θ \equiv sin^{2} θ

Proof Strategy

Step-by-Step Approach

Choose the more complex side — usually has more terms or fractions

Convert to sin and cos — often simplifies the expression

Look for identities — can you substitute sin²θ + cos²θ = 1?

Simplify algebraically — factorise, find common denominators, cancel

Keep working until you reach the other side

Common Mistakes

Working on BOTH sides simultaneously (not allowed!)

Using = instead of ≡ for identities

Forgetting to state which side you're working from

Skipping steps — every line must follow logically

Worked Example — Quotient Identity01

Prove that $\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ} \equiv \frac{1}{s i n θ c o s θ}$

LHS:

\frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}

Find common denominator:

= \frac{s i n ^{2} θ}{s i n θ c o s θ} + \frac{c o s ^{2} θ}{s i n θ c o s θ}

Combine fractions:

= \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ}

Use Pythagorean identity:

sin^{2} θ + cos^{2} θ = 1

= \frac{1}{s i n θ c o s θ}

≡ RHS ✓

Worked Example — Pythagorean Identity02

Prove that $\frac{1 - c o s ^{2} θ}{c o s ^{2} θ} \equiv tan^{2} θ$

LHS:

\frac{1 - c o s ^{2} θ}{c o s ^{2} θ}

Use

1 - cos^{2} θ = sin^{2} θ

:

= \frac{s i n ^{2} θ}{c o s ^{2} θ}

Use

\frac{s i n θ}{c o s θ} = tan θ

:

= (\frac{s i n θ}{c o s θ})^{2} = tan^{2} θ

≡ RHS ✓

Worked Example — Common Denominator03

Prove that $\frac{1}{1 - s i n θ} + \frac{1}{1 + s i n θ} \equiv 2 sec^{2} θ$

LHS:

\frac{1}{1 - s i n θ} + \frac{1}{1 + s i n θ}

Common denominator:

(1 - sin θ) (1 + sin θ)

= \frac{1 + s i n θ + 1 - s i n θ}{( 1 - s i n θ ) ( 1 + s i n θ )}

Simplify numerator and use difference of squares:

= \frac{2}{1 - s i n ^{2} θ}

Use

1 - sin^{2} θ = cos^{2} θ

:

= \frac{2}{c o s ^{2} θ} = 2 sec^{2} θ

≡ RHS ✓

Worked Example — Difference of Squares04

Prove that $\frac{c o s ^{2} θ - s i n ^{2} θ}{c o s θ - s i n θ} \equiv cos θ + sin θ$

LHS:

\frac{c o s ^{2} θ - s i n ^{2} θ}{c o s θ - s i n θ}

Factorise numerator using difference of squares:

= \frac{( c o s θ + s i n θ ) ( c o s θ - s i n θ )}{c o s θ - s i n θ}

Cancel common factor

(cos θ - sin θ)

:

= cos θ + sin θ

≡ RHS ✓

Using Double Angle Formulae

Double angle formulae are often needed in proofs. Remember all forms:

Double Angle Formulae

Sine:

sin 2 θ \equiv 2 sin θ cos θ

Tangent:

tan 2 θ \equiv \frac{2 t a n θ}{1 - t a n ^{2} θ}

Cosine (3 forms):

cos 2 θ \equiv cos^{2} θ - sin^{2} θ

cos 2 θ \equiv 2 cos^{2} θ - 1

cos 2 θ \equiv 1 - 2 sin^{2} θ

Worked Example — Double Angle05

Prove that $\frac{s i n 2 θ}{1 + c o s 2 θ} \equiv tan θ$

LHS:

\frac{s i n 2 θ}{1 + c o s 2 θ}

Use

sin 2 θ = 2 sin θ cos θ

and

cos 2 θ = 2 cos^{2} θ - 1

:

= \frac{2 s i n θ c o s θ}{1 + 2 c o s ^{2} θ - 1}

Simplify denominator:

= \frac{2 s i n θ c o s θ}{2 c o s ^{2} θ}

Cancel common factor

2 cos θ

:

= \frac{s i n θ}{c o s θ} = tan θ

≡ RHS ✓

Worked Example — Multi-Step Proof06

Prove that $\frac{s e c θ - 1}{s e c θ + 1} \equiv \frac{1 - c o s θ}{1 + c o s θ}$

LHS:

\frac{s e c θ - 1}{s e c θ + 1}

Replace

sec θ

with

\frac{1}{c o s θ}

:

= \frac{\frac{1}{c o s θ} - 1}{\frac{1}{c o s θ} + 1}

Multiply numerator and denominator by

cos θ

:

= \frac{1 - c o s θ}{1 + c o s θ}

≡ RHS ✓

Worked Example — Compound Expression07

Prove that $tan θ + cot θ \equiv sec θ csc θ$

LHS:

tan θ + cot θ

Convert to sin and cos:

= \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}

Find common denominator:

= \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ}

Use Pythagorean identity:

= \frac{1}{s i n θ c o s θ}

Rewrite as product of reciprocals:

= \frac{1}{c o s θ} \cdot \frac{1}{s i n θ} = sec θ csc θ

≡ RHS ✓

Practice

Prove that $\frac{sin θ}{1 + cos θ} + \frac{1 + cos θ}{sin θ} \equiv 2 csc θ$

LHS:

\frac{s i n θ}{1 + c o s θ} + \frac{1 + c o s θ}{s i n θ}

Find common denominator

sin θ (1 + cos θ)

:

= \frac{s i n ^{2} θ + ( 1 + c o s θ ) ^{2}}{s i n θ ( 1 + c o s θ )}

Expand:

sin^{2} θ + 1 + 2 cos θ + cos^{2} θ = 2 + 2 cos θ

= \frac{2 ( 1 + c o s θ )}{s i n θ ( 1 + c o s θ )} = \frac{2}{s i n θ} = 2 csc θ

≡ RHS ✓

Practice

Prove that $\frac{1 - cos 2 θ}{sin 2 θ} \equiv tan θ$

LHS:

\frac{1 - c o s 2 θ}{s i n 2 θ}

Use

cos 2 θ = 1 - 2 sin^{2} θ

and

sin 2 θ = 2 sin θ cos θ

:

= \frac{1 - ( 1 - 2 s i n ^{2} θ )}{2 s i n θ c o s θ}

Simplify numerator:

= \frac{2 s i n ^{2} θ}{2 s i n θ c o s θ}

Cancel

2 sin θ

:

= \frac{s i n θ}{c o s θ} = tan θ

≡ RHS ✓

Practice

Prove that $sec^{4} θ - tan^{4} θ \equiv sec^{2} θ + tan^{2} θ$

LHS:

sec^{4} θ - tan^{4} θ

Recognise as difference of squares

a^{2} - b^{2} = (a + b) (a - b)

:

= (sec^{2} θ + tan^{2} θ) (sec^{2} θ - tan^{2} θ)

Use identity

sec^{2} θ - tan^{2} θ = 1

:

= (sec^{2} θ + tan^{2} θ) (1)

= sec^{2} θ + tan^{2} θ

≡ RHS ✓

Summary

Key Proof Strategies

StrategyWhen to Use

Convert to sin and cosWhen you see tan, sec, csc, cot

Common denominatorAdding/subtracting fractions

FactoriseDifference of squares, common factors

Pythagorean identityWhen you see sin² + cos², 1 - sin², etc.

Double angle formulaeWhen you see sin 2θ, cos 2θ, tan 2θ

Remember

Work on ONE side only — transform it to match the other

Use the ≡ symbol for identities

Show ALL steps clearly — marks are awarded for working

Choose the more complex side to simplify