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Radians, arcs and small angles questions
A radian measures an angle by the arc length it cuts on a unit circle, which is the measure the rest of the course needs. In radians the arc length and sector area formulae become s = rθ and A = ½r²θ, the derivatives of sine and cosine take their standard form, and for small angles the trig functions are well approximated by simple polynomials.
7 original questions · 21 marks · the radians, arcs and small angles notes · Trigonometry
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Convert 60° to radians and 3π/4 radians to degrees, exactly.
Worked answer
Since 180° = π, dividing by 3 gives 60° = π/3. Going the other way, 3π/4 = 3 × 45° = 135°. B1 B1 for the two conversions. Multiples of π over small denominators cover every angle the exact-value table knows. Exactly means the π stays; 1.047 is not an answer here.A sector has radius 5 cm and angle 0.8 radians. Find its arc length and area.
Worked answer
The arc is s = rθ = 5 × 0.8 = 4 cm, and the area is ½r2θ = ½ × 25 × 0.8 = 10 cm2. B1 for the arc length, B1 for the area. No factor of 360 appears anywhere, which is the whole reward for working in radians. Both formulae are only valid with θ in radians, so an angle handed over in degrees must be converted first.A sector has radius 9 cm and angle 1.4 radians. Find its perimeter and its area.
Worked answer
The arc is 9 × 1.4 = 12.6 cm, and the perimeter adds the two radii: 12.6 + 18 = 30.6 cm. The area is ½ × 81 × 1.4 = 56.7 cm2. B1 for the arc, B1 for adding the two radii, B1 for the area. Forgetting the two straight edges in the perimeter is the standard slip.An arc of length 12 cm subtends an angle θ at the centre of a circle of radius 7.5 cm. Find θ, and the area of the sector.
Worked answer
Rearranging s = rθ gives θ = 12/7.5 = 1.6 radians, and then the area is ½ × 7.52 × 1.6 = 45 cm2. M1 for rearranging s = rθ, A1 for 1.6 radians, B1 for the area. The formulae run backwards as happily as forwards, and the θ that comes out is already in radians, so it drops straight into the area formula. Converting it to degrees on the way is wasted work and usually loses accuracy.Find the area of the minor segment cut off by a chord in a circle of radius 6 cm, where the chord subtends 1.2 radians at the centre. Give 3 significant figures.
Worked answer
A segment is the sector with the triangle taken out. ½ × 36 × 1.2 − ½ × 36 × sin 1.2 = 21.6 − 16.78 = 4.82 cm2. The triangle uses ½r2 sin θ, with the two radii as its sides and θ between them. M1 for sector minus triangle, A1 for the two areas, A1 for 4.82 cm2. Two things sink this question. The calculator must be in radian mode, since sin 1.2 in degrees is 0.0209 and the answer collapses; and the two terms are close together, so carry full accuracy into the subtraction rather than rounding each part first.Using the small-angle approximations, find the approximate value of (1 − cos 2θ)/(θ sin θ) when θ is small and in radians.
Worked answer
cos 2θ ≈ 1 − (2θ)2/2 = 1 − 2θ2, so the top is approximately 2θ2. The bottom is θ × sin θ ≈ θ × θ = θ2. The ratio is 2θ2/θ2 = 2, independent of θ. M1 for the expansion of cos 2θ, A1 for the numerator, M1 for sin θ ≈ θ below, A1 for the value 2. The replacement for cos must use the angle actually present, 2θ, squared whole: (2θ)2 = 4θ2, not 2θ2.The approximation sin θ ≈ θ is stated for θ in radians. Show it performs well at θ = 0.1 radians, and explain why the same statement fails completely if θ is measured in degrees.
Worked answer
sin 0.1 = 0.0998…, within a fifth of a percent of 0.1. In radians a small angle's arc and sine are nearly the same length, which is what the approximation says. In degrees the claim reads sin 0.1° ≈ 0.1, but sin 0.1° = 0.0017…: the number 0.1 no longer measures the arc, so the geometry behind the approximation is gone. B1 for sin 0.1 = 0.0998, B1 for why arc and sine nearly agree, B1 for sin 0.1° = 0.0017, B1 for why the degree version fails.
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