Exercise 3.2 answers: Trigonometric Functions

Open NCERT p. 57Matches NCERT’s answer

1. Use sin²x + cos²x = 1: sin²x = 1 − 1/4 = 3/4, so sin x = ±√3/2.
2. In the third quadrant sine is negative, so sin x = −√3/2.
3. tan x = sin x ÷ cos x = (−√3/2) ÷ (−1/2) = √3.
4. cot x = 1/tan x = 1/√3 = √3/3.
5. sec x = 1/cos x = −2.
6. cosec x = 1/sin x = −2/√3 = −2√3/3.

Answersin x = −√3/2, tan x = √3, cot x = √3/3, sec x = −2, cosec x = −2√3/3

Watch this explained “Backwards, from one value”, 10:56 into Which of the six stays positive where, and what happens at the quarter turns

Open NCERT p. 57Matches NCERT’s answer

1. Use sin²x + cos²x = 1: cos²x = 1 − 9/25 = 16/25, so cos x = ±4/5.
2. In the second quadrant cosine is negative, so cos x = −4/5.
3. tan x = sin x ÷ cos x = (3/5) ÷ (−4/5) = −3/4.
4. cot x = 1/tan x = −4/3.
5. sec x = 1/cos x = −5/4.
6. cosec x = 1/sin x = 5/3.

Answercos x = −4/5, tan x = −3/4, cot x = −4/3, sec x = −5/4, cosec x = 5/3

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Open NCERT p. 57Matches NCERT’s answer

1. Use 1 + cot²x = cosec²x: cosec²x = 1 + 9/16 = 25/16, so cosec x = ±5/4.
2. In the third quadrant sine is negative, so cosec x = −5/4, and sin x = 1/cosec x = −4/5.
3. cos x = cot x × sin x = (3/4) × (−4/5) = −3/5.
4. tan x = 1/cot x = 4/3.
5. sec x = 1/cos x = −5/3.

Answersin x = −4/5, cos x = −3/5, tan x = 4/3, sec x = −5/3, cosec x = −5/4

Watch this explained “Backwards, from one value”, 10:56 into Which of the six stays positive where, and what happens at the quarter turns

Open NCERT p. 57Matches NCERT’s answer

1. cos x = 1/sec x = 5/13.
2. Use sin²x + cos²x = 1: sin²x = 1 − 25/169 = 144/169, so sin x = ±12/13.
3. In the fourth quadrant sine is negative, so sin x = −12/13.
4. tan x = sin x ÷ cos x = −12/5.
5. cot x = 1/tan x = −5/12.
6. cosec x = 1/sin x = −13/12.

Answersin x = −12/13, cos x = 5/13, tan x = −12/5, cot x = −5/12, cosec x = −13/12

Watch this explained “Backwards, from one value”, 10:56 into Which of the six stays positive where, and what happens at the quarter turns

Open NCERT p. 57Matches NCERT’s answer

1. Use 1 + tan²x = sec²x: sec²x = 1 + 25/144 = 169/144, so sec x = ±13/12.
2. In the second quadrant cosine is negative, so sec x = −13/12, and cos x = 1/sec x = −12/13.
3. sin x = tan x × cos x = (−5/12) × (−12/13) = 5/13.
4. cot x = 1/tan x = −12/5.
5. cosec x = 1/sin x = 13/5.

Answersin x = 5/13, cos x = −12/13, cot x = −12/5, sec x = −13/12, cosec x = 13/5

Watch this explained “Backwards, from one value”, 10:56 into Which of the six stays positive where, and what happens at the quarter turns

Open NCERT p. 57Matches NCERT’s answer

1. 765° = 2 × 360° + 45°.
2. Sine repeats every 360°, so sin 765° = sin 45°.
3. sin 45° = √2/2.

Answer√2/2

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Open NCERT p. 57Matches NCERT’s answer

1. −1410° + 4 × 360° = −1410° + 1440° = 30°.
2. Cosecant repeats every 360°, so cosec(−1410°) = cosec 30°.
3. cosec 30° = 1/sin 30° = 1/(1/2) = 2.

Answer2

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Open NCERT p. 57Matches NCERT’s answer

1. 19π/3 − 6π = 19π/3 − 18π/3 = π/3.
2. Tangent repeats every 2π (also every π), so tan(19π/3) = tan(π/3).
3. tan(π/3) = √3.

Answer√3

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Open NCERT p. 57Matches NCERT’s answer

1. −11π/3 + 4π = −11π/3 + 12π/3 = π/3.
2. Sine repeats every 2π, so sin(−11π/3) = sin(π/3).
3. sin(π/3) = √3/2.

Answer√3/2

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Open NCERT p. 57Matches NCERT’s answer

1. −15π/4 + 4π = −15π/4 + 16π/4 = π/4.
2. Cotangent repeats every 2π (also every π), so cot(−15π/4) = cot(π/4).
3. cot(π/4) = 1.

Answer1

Watch this explained “Bringing a wild input home”, 12:03 into Where each function is defined, what values it reaches, and how it repeats