RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
In earlier classes trigonometric ratios only made sense inside a right triangle, so angles had to stay between 0° and 90°. Here the same ratios are redefined using a unit circle, which lets θ be any real number — negative, obtuse or several full turns. That freedom is what makes the compound-angle formulas, the identities and the general solutions of trigonometric equations possible.
The degree measure of an angle of 5π/12 radians is:
Answer
The angle is 75°. Since π radians = 180°, one radian equals (180/π)°. So 5π/12 radians = (5π/12) × (180/π)° = (5 × 180)/12 degrees = 900/12 = 75°. The factor of π cancels, which is why the answer is a clean whole number of degrees.
If sin θ = 5/13 and θ lies in the second quadrant, then cos θ equals:
Answer
cos θ = -12/13. From sin²θ + cos²θ = 1, cos²θ = 1 - (5/13)² = 1 - 25/169 = 144/169, so cos θ = ±12/13. In the second quadrant the cosine is negative (only sine and cosecant are positive there). Hence cos θ = -12/13. The value -5/12 is tan θ, not cos θ.
The exact value of cos 75° is:
Answer
cos 75° = (√6 - √2)/4. Write 75° = 45° + 30° and use cos(A + B) = cos A cos B - sin A sin B. cos 75° = cos 45° cos 30° - sin 45° sin 30° = (√2/2)(√3/2) - (√2/2)(1/2). This equals √6/4 - √2/4 = (√6 - √2)/4. The first option, (√6 + √2)/4, is cos 15° (equivalently sin 75°).
The general solution of the equation sin x = 0 is:
Answer
The general solution is x = nπ, n ∈ Z. The sine vanishes at 0, π, 2π, -π, -2π and so on, i.e. at every integer multiple of π. Using the rule sin x = sin y giving x = nπ + (-1)ⁿ y with y = 0, the term (-1)ⁿ × 0 = 0, so x = nπ. The option x = 2nπ misses the odd multiples such as π, and x = (2n + 1)π/2 is the solution set of cos x = 0.
Assertion (A): sin 150° = 1/2. Reason (R): For any angle θ, sin(180° - θ) = sin θ.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is a standard allied-angle identity: an angle and its supplement have the same sine, because both terminal arms give the same height on the unit circle. Applying it, sin 150° = sin(180° - 30°) = sin 30° = 1/2, so Assertion A is true. Since the identity in R is exactly the step used to evaluate sin 150°, R correctly explains A.
Find the length of the arc of a circle of radius 18 cm that subtends an angle of 5π/6 radians at the centre. Leave your answer in terms of π and also give it correct to one decimal place.
Answer
The arc length formula is l = rθ, with θ measured in radians. Here r = 18 cm and θ = 5π/6, so l = 18 × 5π/6. Since 18/6 = 3, this gives l = 3 × 5π = 15π cm. Numerically, 15π ≈ 15 × 3.1416 = 47.1 cm correct to one decimal place.
Prove that (1 + tan²θ)/(1 + cot²θ) = tan²θ, stating the values of θ for which the identity is valid.
Answer
Start with the Pythagorean identities 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. So LHS = sec²θ/cosec²θ. Write each in terms of sine and cosine: sec²θ = 1/cos²θ and cosec²θ = 1/sin²θ. Therefore LHS = (1/cos²θ) × (sin²θ/1) = sin²θ/cos²θ = tan²θ = RHS. The identity holds for all θ where every function involved is defined, i.e. cos θ ≠ 0 and sin θ ≠ 0, which means θ ≠ nπ/2 for any integer n.
Evaluate sin 15° exactly, using a compound-angle formula.
Answer
Write 15° = 45° - 30° and use sin(A - B) = sin A cos B - cos A sin B. sin 15° = sin 45° cos 30° - cos 45° sin 30°. Substituting the standard values, = (√2/2)(√3/2) - (√2/2)(1/2) = √6/4 - √2/4. Hence sin 15° = (√6 - √2)/4, which is approximately 0.2588.
If tan θ = 3/4 and θ lies in the third quadrant, find sin θ, cos θ, sin 2θ, cos 2θ and tan 2θ, showing all working.
Answer
Step 1 — Since tan θ = 3/4, the reference right triangle has opposite 3 and adjacent 4, so the hypotenuse is √(3² + 4²) = √25 = 5. Step 2 — In the third quadrant only tangent and cotangent are positive, so both sine and cosine are negative. Hence sin θ = -3/5 and cos θ = -4/5. (Check: (-3/5)² + (-4/5)² = 9/25 + 16/25 = 1.) Step 3 — sin 2θ = 2 sin θ cos θ = 2 × (-3/5) × (-4/5) = 24/25. Step 4 — cos 2θ = cos²θ - sin²θ = 16/25 - 9/25 = 7/25. Step 5 — tan 2θ = sin 2θ/cos 2θ = (24/25)/(7/25) = 24/7. As a check, the double-angle formula gives tan 2θ = 2 tan θ/(1 - tan²θ) = (3/2)/(1 - 9/16) = (3/2)/(7/16) = 24/7, which agrees.
(a) Find the general solution of 2 cos²x + 3 sin x = 0. (b) Prove that sin(A + B) sin(A - B) = sin²A - sin²B.
Answer
(a) Replace cos²x by 1 - sin²x: 2(1 - sin²x) + 3 sin x = 0. Expanding, 2 - 2sin²x + 3 sin x = 0, and multiplying by -1 gives 2sin²x - 3 sin x - 2 = 0. Factorise by splitting the middle term: 2sin²x - 4 sin x + sin x - 2 = 2 sin x(sin x - 2) + 1(sin x - 2) = (2 sin x + 1)(sin x - 2). So either sin x = 2 or sin x = -1/2. The value 2 is rejected because -1 ≤ sin x ≤ 1. Now sin x = -1/2 = sin(-π/6), so the general solution is x = nπ + (-1)ⁿ(-π/6), i.e. x = nπ - (-1)ⁿ π/6, n ∈ Z. (b) Expand both factors. sin(A + B) sin(A - B) = (sin A cos B + cos A sin B)(sin A cos B - cos A sin B). This is of the form (p + q)(p - q) = p² - q², so it equals sin²A cos²B - cos²A sin²B. Write cos²B = 1 - sin²B and cos²A = 1 - sin²A: = sin²A(1 - sin²B) - (1 - sin²A)sin²B = sin²A - sin²A sin²B - sin²B + sin²A sin²B. The two middle terms cancel, leaving sin²A - sin²B, as required.
Read the following and answer the questions that follow: A giant wheel at a fairground has radius 25 m and its centre is fixed 30 m above the ground. A seat starts level with the centre and rises as the wheel turns. If θ is the angle (in radians) turned from that starting position, the height of the seat above the ground is modelled by h(θ) = 30 + 25 sin θ metres. (a) Find the greatest and least heights reached by the seat. (b) Find the height when θ = π/6. (c) Find all values of θ in [0, 2π) for which the seat is exactly 30 m above the ground. (d) Find the distance travelled along the rim by the seat when the wheel turns through π/3 radians.
Answer
(a) Since -1 ≤ sin θ ≤ 1, the term 25 sin θ ranges from -25 to 25. Greatest height = 30 + 25 = 55 m (when sin θ = 1); least height = 30 - 25 = 5 m (when sin θ = -1). (b) h(π/6) = 30 + 25 sin(π/6) = 30 + 25 × (1/2) = 30 + 12.5 = 42.5 m. (c) Set 30 + 25 sin θ = 30, so 25 sin θ = 0 and sin θ = 0. In [0, 2π) the sine vanishes at θ = 0 and θ = π. So the seat is level with the centre at those two positions. (d) The rim distance is an arc length, l = rθ = 25 × π/3 = 25π/3 metres ≈ 26.2 m.
RowQ generates fresh questions on Trigonometric Functions, marks your answers, and explains every step.
Start free