Trigonometric ratios and identities for NDA

Trigonometry is the chapter that quietly runs through the rest of the NDA maths paper. Heights and distances, properties of triangles, inverse functions, complex numbers in polar form, and half the integrals you will meet all lean on the same identities. Questions on the chapter itself are usually short: evaluate an expression, simplify, or find a maximum. They reward the candidate who recognises the identity at a glance and punish the one who expands everything.

Typical question types are:

• converting between degrees and radians, or using arc length;
• values at standard and allied angles (150°, 210°, 315° and so on);
• simplifying an expression to a constant with the Pythagorean identities;
• finding one ratio from another, often with a sign decided by the quadrant;
• compound, double and triple angle evaluations such as sin 75° or cos 3A;
• products like sin 20° sin 40° sin 80°;
• the greatest or least value of an expression.

Angles and radians

π radians = 180°, so multiply by π/180 to go from degrees to radians and by 180/π to come back. 1 radian is about 57.3°. An arc of a circle of radius r subtending θ radians at the centre has length l = rθ. The formula needs θ in radians, which is where most slips happen.

Standard values and allied angles

Two more pairs appear often enough to learn: sin 15° = cos 75° = (√3 − 1)/(2√2), tan 15° = 2 − √3 and tan 75° = 2 + √3. Also sin 18° = (√5 − 1)/4 and cos 36° = (√5 + 1)/4.

Signs by quadrant. In the first quadrant all ratios are positive. In the second only sin (and cosec) is positive, in the third only tan (and cot), in the fourth only cos (and sec).

Allied angles. For 180° ± θ and 360° − θ the ratio keeps its name. For 90° ± θ and 270° ± θ it changes to the co-ratio (sin to cos, tan to cot). In both cases the sign is the sign of the original ratio in the quadrant where the angle lands. So cos 210° = cos(180° + 30°) = −cos 30° = −√3/2, because cos is negative in the third quadrant.

The Pythagorean identities

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

The second and third are the first one divided by cos²θ and by sin²θ. A useful consequence: (sec θ + tan θ)(sec θ − tan θ) = 1, so if you know sec θ + tan θ you know sec θ − tan θ as its reciprocal. The same holds for cosec θ + cot θ.

Compound, double and triple angles

• sin(A ± B) = sin A cos B ± cos A sin B
• cos(A ± B) = cos A cos B ∓ sin A sin B
• tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
• sin(A + B) sin(A − B) = sin²A − sin²B

Put B = A to get the double angles:

• sin 2A = 2 sin A cos A = 2 tan A/(1 + tan²A)
• cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1 = (1 − tan²A)/(1 + tan²A)
• tan 2A = 2 tan A/(1 − tan²A)

Rearranging the cos 2A forms gives sin²A = (1 − cos 2A)/2 and cos²A = (1 + cos 2A)/2, which you will need again in integration.

Triple angles: sin 3A = 3 sin A − 4 sin³A, cos 3A = 4 cos³A − 3 cos A, tan 3A = (3 tan A − tan³A)/(1 − 3 tan²A).

Sum-to-product and product-to-sum

Two product results save a lot of time: sin θ sin(60° − θ) sin(60° + θ) = ¼ sin 3θ and cos θ cos(60° − θ) cos(60° + θ) = ¼ cos 3θ.

Greatest and least values

For any θ, −√(a² + b²) ≤ a sin θ + b cos θ ≤ √(a² + b²). The reason: write a = R cos α and b = R sin α with R = √(a² + b²). Then a sin θ + b cos θ = R sin(θ + α), and a sine lies between −1 and 1.

Other quick bounds: sin θ cos θ lies between −1/2 and 1/2, and for positive p and q the least value of p tan²θ + q cot²θ is 2√(pq) by AM ≥ GM.

Worked NDA-style MCQs

Q1. If sin θ + cos θ = √2, then sin θ cos θ equals:
(a) 1 (b) 1/2 (c) 1/4 (d) √2/2

Square both sides: 1 + 2 sin θ cos θ = 2, so sin θ cos θ = 1/2. As a bonus, tan θ + cot θ = 1/(sin θ cos θ) = 2. Answer: (b).

Q2. tan 15° + cot 15° equals:
(a) 2 (b) 2√3 (c) 4 (d) √3

(2 − √3) + (2 + √3) = 4. Or, faster: tan θ + cot θ = 1/(sin θ cos θ) = 2/sin 2θ = 2/sin 30° = 4. Answer: (c).

Q3. If sec θ + tan θ = 3, then sin θ equals:
(a) 3/5 (b) 4/5 (c) 4/3 (d) 5/4

sec θ − tan θ = 1/3. Adding and subtracting: sec θ = 5/3 and tan θ = 4/3. So sin θ = tan θ/sec θ = (4/3) × (3/5) = 4/5. Answer: (b).

Q4. The greatest value of 3 sin θ − 4 cos θ + 7 is:
(a) 5 (b) 7 (c) 12 (d) 14

3 sin θ − 4 cos θ lies between −5 and 5, since √(9 + 16) = 5. The greatest value of the whole expression is 5 + 7 = 12 (and the least is 2). Answer: (c).

Q5. sin 20° sin 40° sin 80° equals:
(a) 1/8 (b) √3/8 (c) √3/4 (d) 3/8

This is sin θ sin(60° − θ) sin(60° + θ) with θ = 20°, so it equals ¼ sin 60° = √3/8. Answer: (b).

Q6. If A + B = 45°, then (1 + tan A)(1 + tan B) equals:
(a) 1 (b) 2 (c) 3 (d) depends on A

tan(A + B) = 1 gives tan A + tan B = 1 − tan A tan B. Expanding the product: 1 + (tan A + tan B) + tan A tan B = 1 + 1 = 2. Answer: (b).

Common mistakes

• Using degrees in l = rθ. Convert to radians first.
• Getting the sign wrong for allied angles. Decide the quadrant, then the sign, then the ratio.
• Writing sin(A + B) = sin A + sin B. No trigonometric function distributes over addition.
• Taking the positive root automatically. If θ is in the second quadrant, cos θ = −√(1 − sin²θ).
• Forgetting that the maximum formula needs the same angle. 3 sin θ + 4 cos 2θ does not have maximum 5.

Practice set

1. 150° in radians is: (a) 5π/6 (b) 3π/4 (c) 2π/3 (d) 7π/6
2. cos 210° equals: (a) √3/2 (b) −1/2 (c) −√3/2 (d) 1/2
3. If tan A = 3/4 and A is acute, sin 2A equals: (a) 12/25 (b) 24/25 (c) 7/25 (d) 3/5
4. If cos A = 1/3, cos 3A equals: (a) −23/27 (b) 23/27 (c) −1 (d) 1/9
5. (sin θ + cosec θ)² + (cos θ + sec θ)² equals: (a) 9 (b) 7 + tan²θ + cot²θ (c) 5 + tan²θ (d) 7
6. The least value of 9 tan²θ + 4 cot²θ is: (a) 13 (b) 6 (c) 12 (d) 36
7. sin 75° + sin 15° equals: (a) √3/2 (b) √2 (c) √6/2 (d) 1
8. tan 15° equals: (a) 2 + √3 (b) 2 − √3 (c) √3 − 1 (d) 1/√3

Answers:

1. (a). 150 × π/180 = 5π/6.
2. (c). Third quadrant, cos negative: −cos 30°.
3. (b). 2 tan A/(1 + tan²A) = (3/2) ÷ (25/16) = 24/25.
4. (a). 4(1/27) − 3(1/3) = 4/27 − 1 = −23/27.
5. (b). Expanding gives sin²θ + cos²θ + cosec²θ + sec²θ + 4 = 1 + (1 + cot²θ) + (1 + tan²θ) + 4.
6. (c). AM ≥ GM: 2√(9 × 4) = 12, reached when tan²θ = 2/3.
7. (c). 2 sin 45° cos 30° = 2 × (1/√2) × (√3/2) = √6/2.
8. (b). tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3.

What to do next

• Write the allied-angle rule and the sum-to-product table from memory, then check them against this page.
• Solve 30 old NDA questions on this chapter, timed at 60–75 seconds each, and note which identity each one needed.
• Apply the ratios straight away in heights and distances, then move on to inverse trigonometric functions and the properties of triangles.