Trigonometry for CDS

Trigonometry in the CDS maths paper is almost entirely about acute angles: ratios in a right triangle, the values at 0°, 30°, 45°, 60° and 90°, a handful of identities, and heights and distances. There is no heavy formula work with compound angles in most questions. What the paper tests is whether you can simplify an expression in two or three lines instead of eight.

That makes this a scoring topic. Learn the table, understand where the identities come from, and practise the "pair" tricks below, and most questions here become 40-second questions.

The six ratios

Take a right triangle with an acute angle θ. Call the side opposite θ the perpendicular (P), the side next to it the base (B), and the longest side the hypotenuse (H).

Because H is the longest side, sin θ and cos θ lie between 0 and 1 for angles from 0° to 90°, while sec θ and cosec θ are at least 1. Any option that says sin θ = 5/4 is impossible, and you can strike it out at once.

Standard values

A quick way to rebuild the sine row: write √0/2, √1/2, √2/2, √3/2, √4/2. That gives 0, 1/2, 1/√2, √3/2, 1. The cosine row is the same list read backwards. As θ rises from 0° to 90°, sin θ and tan θ increase, and cos θ decreases.

The three identities, and why they hold

In the right triangle, P² + B² = H². Divide the whole equation by H², then by B², then by P²:

• Divide by H²: sin²θ + cos²θ = 1
• Divide by B²: tan²θ + 1 = sec²θ, so sec²θ − tan²θ = 1
• Divide by P²: 1 + cot²θ = cosec²θ, so cosec²θ − cot²θ = 1

So there is only one fact here, Pythagoras, written three ways. If you forget an identity in the exam hall, rebuild it in ten seconds this way.

Pair tricks

The second and third identities factorise:

• (sec θ + tan θ)(sec θ − tan θ) = 1. So if sec θ + tan θ = p, then sec θ − tan θ = 1/p.
• (cosec θ + cot θ)(cosec θ − cot θ) = 1. So if cosec θ − cot θ = q, then cosec θ + cot θ = 1/q.
• If sin θ + cos θ = k, squaring gives 1 + 2 sin θ cos θ = k², so sin θ cos θ = (k² − 1)/2.

Adding and subtracting the pair gives sec θ and tan θ separately. This is one of the most common CDS question types.

Complementary angles

In a right triangle the two acute angles add to 90°. The side opposite one is the side next to the other, so:

• sin(90° − θ) = cos θ and cos(90° − θ) = sin θ
• tan(90° − θ) = cot θ and cot(90° − θ) = tan θ
• sec(90° − θ) = cosec θ and cosec(90° − θ) = sec θ

Two consequences appear again and again: tan θ × tan(90° − θ) = 1, and sin²θ + sin²(90° − θ) = 1.

Maximum and minimum values

• For any θ, a sin θ + b cos θ lies between −√(a² + b²) and √(a² + b²). So the maximum of 3 sin θ + 4 cos θ is 5.
• If two positive terms multiply to a constant, their sum is least when they are equal (AM ≥ GM). So a tan²θ + b cot²θ ≥ 2√(ab), since tan θ × cot θ = 1.

Worked questions

Question 1: If tan θ = 8/15 and θ is acute, find sin θ + cos θ.

• P = 8, B = 15, so H = √(64 + 225) = √289 = 17.
• sin θ + cos θ = 8/17 + 15/17 = 23/17.

Question 2: Find the value of 4 sin²30° + 3 tan²60° − 2 cos²45°.

• 4 × 1/4 + 3 × 3 − 2 × 1/2 = 1 + 9 − 1 = 9.

Question 3: If sec θ + tan θ = 2, find sin θ.

• sec θ − tan θ = 1/2.
• Adding: 2 sec θ = 5/2, so sec θ = 5/4. Subtracting: 2 tan θ = 3/2, so tan θ = 3/4.
• sin θ = tan θ ÷ sec θ = (3/4) × (4/5) = 3/5.

Question 4: If sin θ + cos θ = √2, find tan θ + cot θ.

• sin θ cos θ = (2 − 1)/2 = 1/2.
• tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin²θ + cos²θ) ÷ (sin θ cos θ) = 1 ÷ (1/2) = 2. (Here θ = 45°, which confirms it.)

Question 5: Find tan 1° × tan 2° × tan 3° × … × tan 89°.

• Pair tan 1° with tan 89°, tan 2° with tan 88°, and so on. Each pair is tan θ × cot θ = 1.
• The middle term is tan 45° = 1. So the product is 1.

Question 6: If sin(A + B) = 1 and cos(A − B) = √3/2, where A and B are acute and A > B, find A and B.

• sin(A + B) = 1 gives A + B = 90°. cos(A − B) = √3/2 gives A − B = 30°.
• So A = 60°, B = 30°.

Simplification habits

• Convert everything to sin and cos when an expression looks messy. Most of it cancels.
• (1 − sin θ)(1 + sin θ) = cos²θ, and (sec θ − 1)(sec θ + 1) = tan²θ. Spot these difference-of-squares forms.
• If the answer must be the same for every θ, put θ = 45° (or 30°) and test the options. Check with a second angle if two options tie.

Practice set

1. If cos θ = 12/13 and θ is acute, find tan θ.
2. Find sin 35°/cos 55° + cos 35°/sin 55°.
3. Simplify (sec²θ − 1)(cosec²θ − 1).
4. If cosec θ − cot θ = 1/3, find cos θ.
5. Find sin²1° + sin²2° + … + sin²89°.
6. Find the value of 2 tan 30° ÷ (1 + tan²30°).
7. If sin θ = cos θ and θ is acute, find 2 tan²θ + sin²θ − 1.
8. Find the minimum value of 9 tan²θ + 4 cot²θ.

Answers

1. 5/12. B = 12, H = 13, so P = 5.
2. 2. Each term is 1, because sin 35° = cos 55° and cos 35° = sin 55°.
3. 1. It equals tan²θ × cot²θ.
4. 4/5. cosec θ + cot θ = 3. Adding gives cosec θ = 5/3, so sin θ = 3/5 and cos θ = 4/5.
5. 44.5. 44 complementary pairs each give 1, and sin²45° = 1/2.
6. √3/2. (2/√3) ÷ (4/3) = 3/(2√3) = √3/2, which is sin 60°.
7. 3/2. θ = 45°, so 2 × 1 + 1/2 − 1.
8. 12. The product of the terms is 36, so the sum is at least 2√36.

What to do next

• Write the full six-row value table from memory, twice, without looking
• Rebuild the three identities from Pythagoras once a day for a week
• Move on to heights and distances, which uses the same table on real figures
• Revise algebraic identities, because many trigonometry simplifications are algebra in disguise
• Solve every trigonometry question from the last five CDS papers, and time each one