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Most students know sin 30° and cos 45° without thinking. Fewer know sin 18°, and JEE questions that use it look impossible until you recognise the angle. The good news is that sin 18° comes from one trick and two identities you already have. This post derives it, gets cos 36° and its relatives, and works through four questions where the values do the heavy lifting.
Is this in the syllabus?
The Syllabus for JEE (Main) 2026 lists "Trigonometrical identities and trigonometrical functions" in Unit 14. The tools for this derivation are in the 2026-27 NCERT Class 11 chapter Trigonometric Functions, which proves sin 3x = 3 sin x − 4 sin³x and cos 3x = 4 cos³x − 3 cos x. The values themselves are not a separate syllabus line, but NCERT's historical note at the end of the chapter says the exact expressions for the sines and cosines of 18°, 36°, 54° and 72° were given by Bhaskara II. Check the syllabus for your own exam year.
How do you derive sin 18°?
Let θ = 18°. Then 5θ = 90°, so 2θ = 90° − 3θ.
Take sine of both sides: sin 2θ = sin(90° − 3θ) = cos 3θ.
Expand: 2 sin θ cos θ = 4 cos³θ − 3 cos θ.
cos 18° is not zero, so divide by cos θ: 2 sin θ = 4 cos²θ − 3.
Replace cos²θ with 1 − sin²θ: 2 sin θ = 1 − 4 sin²θ.
With s = sin θ, this is 4s² + 2s − 1 = 0, so s = (−2 ± √20)/8 = (−1 ± √5)/4.
sin 18° is positive, so sin 18° = (√5 − 1)/4 ≈ 0.309.
Step 6 is where marks are lost. The other root, (−1 − √5)/4 ≈ −0.809, is a real solution of the quadratic, but it is the sine of a different angle (−54°, as it happens). You must reject it with a reason.
How do you get cos 36° from it?
Use the double-angle formula cos 2θ = 1 − 2 sin²θ with θ = 18°:
sin²18° = (6 − 2√5)/16
cos 36° = 1 − (6 − 2√5)/8 = (2 + 2√5)/8 = (√5 + 1)/4 ≈ 0.809
A second route checks the first. Let θ = 36°. Then 2θ = 180° − 3θ, so sin 2θ = sin 3θ. Divide by sin θ: 2 cos θ = 3 − 4 sin²θ = 4 cos²θ − 1. So 4c² − 2c − 1 = 0 and the positive root is (1 + √5)/4. Same answer.
What about sin 36°, cos 18° and the rest?
Use co-function and Pythagorean identities, taking positive roots because all these angles are acute:
cos 72° = sin 18° = (√5 − 1)/4
sin 54° = cos 36° = (√5 + 1)/4
cos 18° = √(1 − sin²18°) = √(10 + 2√5)/4 ≈ 0.951
sin 36° = √(1 − cos²36°) = √(10 − 2√5)/4 ≈ 0.588
There is also a link to geometry. (1 + √5)/2 is the golden ratio φ ≈ 1.618, so cos 36° = φ/2. In a regular pentagon, each interior angle is 108°, and the triangle formed by two sides and a diagonal has angles 108°, 36° and 36°. Drop a perpendicular from the 108° vertex: half the diagonal is (side) × cos 36°, so the diagonal is 2 cos 36° = φ times the side.
Worked examples
1. Find cos 36° − cos 72°. (√5 + 1)/4 − (√5 − 1)/4 = 2/4 = 1/2.
2. Find sin 18° × cos 36°. [(√5 − 1)(√5 + 1)]/16 = (5 − 1)/16 = 1/4. Spot the difference of squares and the surds disappear.
3. Show that sin²72° − sin²60° = (√5 − 1)/8. sin²72° = cos²18° = (10 + 2√5)/16, and sin²60° = 12/16. The difference is (2√5 − 2)/16 = (√5 − 1)/8, which is half of sin 18°.
4. Find cos 36° cos 72° cos 108° cos 144°. cos 108° = −cos 72° and cos 144° = −cos 36°, so the product is (cos 36° cos 72°)². From example 2 that is (1/4)² = 1/16.
We checked every value here numerically against a calculator. In the exam you do not have one, so check signs and sizes instead: a cosine of an acute angle must be between 0 and 1.
Which mistakes cost the most marks?
Keeping the negative root. sin 18° must be positive.
Dividing by cos θ without saying why. It is allowed only because cos 18° ≠ 0. In a proof, write that.
Mixing up the triple-angle formulas. sin 3x has 3 sin x first; cos 3x has 4 cos³x first.
Memorising surds without the method. If you forget whether it is √5 − 1 or √5 + 1, the size check (0.309 or 0.809) settles it.
Writing cos 18° as (√5 + 1)/4. That is cos 36°. cos 18° needs a nested root.
A practice routine
Studying a worked derivation helps most when you explain each step to yourself. In Chi and colleagues (1989), students who learned well from worked mechanics examples generated many self-explanations and related each step to principles in the text; weaker students did not, and leaned on the examples later.
Day 1: Derive sin 18° from a blank page, saying aloud why each line follows.
Day 2: Derive cos 36° both ways and check they agree.
Day 3: Build the full list: sin and cos of 18°, 36°, 54° and 72°.
Day 4: Five product and sum questions like the examples above, without notes.
For the identities this depends on, see our post on practising trigonometric identities. If a concept underneath is shaky, such as why cos(90° − x) = sin x, the Fermi Concept Library covers every concept in mathematics, each with its own lesson.
Tomorrow morning, derive sin 18° in under three minutes with your book closed.
