Fundamental trig identities and using them to simplify expressions and solve equations.
Answer at least 3 of 3 correctly to complete this section.
At GCSE you used trigonometry to find sides and angles in triangles. At A-Level, trigonometry becomes a language for describing waves, rotations, and periodic phenomena. The identities you learn here are not just algebra tricks — they are tools you will use in differentiation, integration, and modelling throughout the course.
The Pythagorean identity sin2θ+cos2θ=1 is the single most important equation in this topic. Almost every trig simplification or proof comes back to it.
These exact values appear in almost every A-Level paper. You need to recall them instantly.
| Angle | sinθ | cosθ | tanθ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | 21 | 23 | 31 |
| 45° | 22 | 22 | 1 |
| 60° | 23 | 21 | 3 |
| 90° | 1 | 0 | undefined |
Memory trick: For sine, the numerators go 0,1,2,3,2 — all over 2. For cosine, it is the same sequence in reverse.
Find the exact value of tan45°+cos30°.
tan45°+cos30°=1+23
This cannot be simplified further. Leave it in this form.
Be careful with fractions. 1+23=21+3. The 1 is not over 2.
Find the exact value of 2sin45°cos45°.
2×22×22=2×42=2×21=1
You might recognise this as sin(2×45°)=sin90°=1 — a preview of the double angle formulae.
Answer at least 3 of 3 correctly to complete this section.
For any angle θ:
sin2θ+cos2θ=1
This comes directly from the unit circle: a point at angle θ on the unit circle has coordinates (cosθ,sinθ), and since the circle has radius 1, Pythagoras gives cos2θ+sin2θ=1.
From this one identity, you can derive two useful rearrangements:
sin2θ=1−cos2θandcos2θ=1−sin2θ
And by dividing through by cos2θ:
cos2θsin2θ+1=cos2θ1⟹tan2θ+1=sec2θ
Given that sinθ=53 and θ is acute, find cosθ.
Using sin2θ+cos2θ=1:
cos2θ=1−259=2516
cosθ=54
(Positive because θ is acute — cosine is positive in the first quadrant.)
Simplify cos2θ1−cos2θ.
Recognise that 1−cos2θ=sin2θ:
cos2θ1−cos2θ=cos2θsin2θ=tan2θ
Given that tanθ=247 and θ is acute, find sinθ.
Using 1+tan2θ=sec2θ:
sec2θ=1+57649=576625
cos2θ=625576⟹cosθ=2524
Then sinθ=tanθ×cosθ=247×2524=257.
Alternative method: You could also draw a right-angled triangle with opposite = 7 and adjacent = 24, then use Pythagoras to find the hypotenuse = 25.
Answer at least 3 of 4 correctly to complete this section.
Often, a trig equation involves more than one function (e.g. both sinθ and cosθ). You use the identities from Section 2 to reduce the equation to a single trigonometric function, then solve using the CAST diagram.
For a full treatment of solving trig equations — including the CAST diagram, multiple angles, and quadratic trig equations — see the Solving Trig Equations lesson.
Solve 2sin2θ−sinθ−1=0 for 0≤θ≤360°.
Let s=sinθ:
2s2−s−1=0⟹(2s+1)(s−1)=0
s=−21ors=1
When sinθ=1: θ=90°
When sinθ=−21: θ=210° or θ=330°
(Sine is negative in the 3rd and 4th quadrants; reference angle is 30°.)
Final answer: θ=90°,210°,330°.
Answer at least 2 of 2 correctly to complete this section.
To prove that the left-hand side (LHS) equals the right-hand side (RHS), you pick one side and manipulate it until it looks like the other side. You never move terms across the equals sign.
Useful strategies:
Prove that 1−cosθ1+1+cosθ1=sin2θ2.
Start with the LHS. Combine the fractions:
LHS=(1−cosθ)(1+cosθ)(1+cosθ)+(1−cosθ)
Numerator: 1+cosθ+1−cosθ=2.
Denominator: (1−cosθ)(1+cosθ)=1−cos2θ=sin2θ.
=sin2θ2=RHS□
Prove that 1−sinθcosθ−1+sinθcosθ=2tanθ.
Start with the LHS:
LHS=(1−sinθ)(1+sinθ)cosθ(1+sinθ)−cosθ(1−sinθ)
Numerator: cosθ+cosθsinθ−cosθ+cosθsinθ=2cosθsinθ.
Denominator: 1−sin2θ=cos2θ.
=cos2θ2cosθsinθ=cosθ2sinθ=2tanθ=RHS□
Exam technique: Always write ”= RHS” (or "□") at the end of a proof to show you have reached the target expression. Never start with “LHS = RHS” — that is assuming what you are trying to prove.
Prove that cosθ1−sin2θ=cosθ.
The first two lines are done for you. Fill in the blanks to complete the proof.
LHS=cosθ1−sin2θ
=cosθcos2θ(use sin2θ+cos2θ=1)
=cosθ=RHS□
Check your answer: The first blank is cos2θ (since 1−sin2θ=cos2θ). The second blank is cosθ (since cosθcos2θ=cosθ).
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Graphs of sin, cos, tan and their transformations. Exact values of trig ratios.
Solving trig equations in given intervals, finding principal and secondary solutions.
Definitions, graphs, and identities involving sec, cosec, and cot.
Addition formulae for sin(A±B), cos(A±B), tan(A±B). Double angle formulae and their applications.
Derivatives of sin x, cos x, tan x and their compositions.