Applications: Radians and Trigonometric Identities

An angle in radians is the arc length divided by the radius, so the crust and the radius give the angle at once. With the angle in radians, the area of a sector is 12r²θ, and no factor of 360 is needed.

1. Let the angle at the center be θ radians. An arc of a circle of radius r that subtends θ radians has length s = rθ, so here 10 = 15θ.
2. (a) Divide both sides by 15: θ = 1015 = 23 radian. In degrees this is 23 × 180^°π = 120^°π, which is 38.2^° to 1 decimal place.
3. The area of a sector with the angle in radians is A = 12r²θ, so A = 12 × 15² × 23.
4. (b) A = 12 × 225 × 23 = 75 cm². Check: the slice is 23 ÷ 2π = 13π of the whole pizza, and 13π × 225π = 75.

answer(a) 23 radian, which is 38.2^° to 1 decimal place; (b) 75 cm²

techniqueArc Length and Sector Area · Radians

examsO-Level · SAT

The bob moves on an arc of a circle whose radius is the rod. A swing from one side to the other turns the rod through π6 on each side of the vertical, and with the angle in radians the arc length is the radius times the angle. The height comes from the right-angled triangle between the rod at the end of the swing and the vertical.

1. From one side to the other the rod turns through π6 + π6 = π3 radian, and the bob moves on a circle of radius 80 cm about the pivot.
2. (a) The arc length is s = rθ = 80 × π3 = 80π3 cm, which is 83.8 cm to 1 decimal place.
3. At the end of a swing the rod makes an angle of π6 with the vertical, so the bob is 80cosπ6 cm below the pivot. The exact value is cosπ6 = √32, so that depth is 40√3 cm.
4. (b) At its lowest point the bob is 80 cm below the pivot, so at the end of a swing it has risen 80 − 40√3 cm, which is 10.7 cm to 1 decimal place. Check: 40√3 ≈ 69.3, and 80 − 69.3 = 10.7.

answer(a) 80π3 cm, which is 83.8 cm; (b) 80 − 40√3 cm, which is 10.7 cm

techniqueArc Length and Sector Area · Exact Trigonometric Values in Radians · Radians

examsO-Level · SAT

The two currents do not peak at the same phase, so their peaks do not simply add. Writing the sum as a single sine wave Rsin(x + α) shows its peak, R, and its shift, α. Expand the compound angle and match the coefficients of sin x and cos x.

1. Expand the compound angle: Rsin(x + α) = Rsin xcosα + Rcos xsinα. For this to equal 3sin x + 4cos x for every x, the coefficients must match: Rcosα = 3 and Rsinα = 4.
2. Square both equations and add: R²(cos²α + sin²α) = 3² + 4² = 25. Since cos²α + sin²α = 1, R² = 25, and R = 5 because R is positive.
3. (a) i = 5sin(x + α). A sine is never more than 1, so the peak current is 5 amps, not the 3 + 4 = 7 amps of adding the two peaks.
4. Divide the second equation by the first: tanα = 43. Both cosα and sinα are positive, so α is acute, and α = tan⁻¹43 = 0.927 radian to 3 significant figures.
5. (b) α = 0.927 radian, about 53.1^°. The combined current is the wave 5sin x moved 0.927 to the left, so it reaches each peak 0.927 radian sooner than 3sin x does. Check: at x = 0, i = 4, and 5sin 0.927 = 5 × 0.8 = 4.

answer(a) i = 5sin(x + α) with tanα = 43, a peak current of 5 amps; (b) α = 0.927 radian, about 53.1^°: the wave 5sin x moved 0.927 to the left

techniqueThe Harmonic Form R sin(x + α) · The Compound Angle Formulas

examsO-Level

Two waves added together make one wave, and in the form Rsin(x + α) its greatest value and its timing can be read off. Here R and α come out exactly, because √3 and 1 are the sides of a 30^°, 60^° right-angled triangle. Each time is then an equation in the single angle x + α.

1. Match Rsin(x + α) = Rsin xcosα + Rcos xsinα with √3sin x + cos x: Rcosα = √3 and Rsinα = 1. Then R² = 3 + 1 = 4, so R = 2, and tanα = 1√3, so α = π6.
2. So h = 2sin(x + π6). The greatest value of a sine is 1, so the greatest height is 2 m, and it comes when x + π6 = π2.
3. (a) x = π2 − π6 = π3: the buoy is first 2 m up after π3 seconds, which is 1.05 seconds.
4. For 1 m below, set h = −1: 2sin(x + π6) = −1, so sin(x + π6) = −12. From x = 0 the angle x + π6 starts at π6, and the first angle after that with a sine of −12 is π + π6 = 7π6.
5. (b) x = 7π6 − π6 = π: the buoy is first 1 m below the calm-water level after π seconds, which is 3.14 seconds. Check: √3sinπ + cosπ = 0 − 1 = −1.

answer(a) h = 2sin(x + π6); the greatest height is 2 m, first after π3 seconds, which is 1.05 s; (b) after π seconds, which is 3.14 s

techniqueSolving a sin x + b cos x = c with the R Form · The Harmonic Form R sin(x + α) · Every Solution of a Trigonometric Equation

examsO-Level

The ball lands when its height is zero again, and the distance across at that moment is the range. The range comes out as a product sinθcosθ, and the double-angle formula turns that product into the single sine sin 2θ, whose greatest value and whose solutions can be read off.

1. The ball lands when its height is 0 again: 20Tsinθ − 5T² = 0, so 5T(4sinθ − T) = 0. T = 0 is the moment of the throw, so the ball lands at T = 4sinθ.
2. The range is the distance across at that time: 20 × 4sinθ × cosθ = 80sinθcosθ. The double-angle formula sin 2θ = 2sinθcosθ turns this into 40sin 2θ m.
3. (a) A sine is at most 1, so the greatest range is 40 m, when 2θ = π2, that is, when θ = π4, or 45^°.
4. For a range of 20 m, 40sin 2θ = 20, so sin 2θ = 12. The ball is thrown upward, so 0 < θ < π2 and 0 < 2θ < π. In that interval 2θ = π6 or 2θ = π − π6 = 5π6.
5. (b) θ = π12 or θ = 5π12, that is, 15^° or 75^°. Check: sin 30^° = sin 150^° = 12, and the two angles add to 90^°, as two throws with the same range must.

answer(a) 40sin 2θ m; the greatest range is 40 m, at θ = π4, which is 45^°; (b) θ = π12 or 5π12, which is 15^° or 75^°

techniqueThe Double Angle Formulas · Exact Trigonometric Values in Radians

examsO-Level

The tip, the pivot and the point straight below the tip make a right-angled triangle with the arm as its hypotenuse, so the height is 4sin 75^° and the horizontal distance 4cos 75^°. There is no exact value for 75^° in the table, but 75^° = 45^° + 30^°, and the compound-angle formulas build its sine and cosine from theirs.

1. The tip is at a height of 4sin 75^° m and a horizontal distance of 4cos 75^° m from the pivot. The angle 75^° is 45^° + 30^°, and both of those have exact values.
2. The compound-angle formula for sine: sin(45^° + 30^°) = sin 45^°cos 30^° + cos 45^°sin 30^° = √22 × √32 + √22 × 12 = √6 + √24.
3. (a) The height is 4 × √6 + √24 = √6 + √2 m, which is 3.86 m to 2 decimal places.
4. The compound-angle formula for cosine has a minus sign: cos(45^° + 30^°) = cos 45^°cos 30^° − sin 45^°sin 30^° = √64 − √24 = √6 − √24.
5. (b) The horizontal distance is √6 − √2 m, which is 1.04 m to 2 decimal places. Check: (√6 + √2)² + (√6 − √2)² = (8 + 2√12) + (8 − 2√12) = 16 = 4², the square of the arm.

answer(a) √6 + √2 m, which is 3.86 m; (b) √6 − √2 m, which is 1.04 m

techniqueThe Compound Angle Formulas · Exact Trigonometric Values in Radians

examsO-Level

In one turn there are two angles with a sine of 12, and a sine repeats every 2π, so the solutions come in two families, each a whole number of turns apart. Solve for the angle first, then turn each family into times, and keep only the times from 0 to 12 seconds.

1. Divide both sides by 8: sinπ x3 = 12. The principal value is π6, and the other angle in one turn with the same sine is π − π6 = 5π6.
2. A sine repeats every 2π, so π x3 = π6 + 2nπ or π x3 = 5π6 + 2nπ, where n is any integer.
3. (a) Multiply every term by 3π: x = 12 + 6n or x = 52 + 6n, where n is an integer. The 6 is the period: 2π × 3π = 6 seconds.
4. Keep the times with 0 ≤ x ≤ 12. n = 0 gives 0.5 and 2.5, and n = 1 gives 6.5 and 8.5. n = 2 gives 12.5 and 14.5, which are too late, and n = −1 gives negative times, which are before the start.
5. (b) The weight passes through the beam at 0.5, 2.5, 6.5 and 8.5 seconds: on the way up at 0.5 and 6.5, and on the way down at 2.5 and 8.5. Check: 8sin6.5π3 = 8sin(2π + π6) = 8 × 12 = 4.

answer(a) x = 12 + 6n or x = 52 + 6n, where n is an integer; (b) at 0.5, 2.5, 6.5 and 8.5 seconds

techniqueEvery Solution of a Trigonometric Equation · Exact Trigonometric Values in Radians

examsSAT

The pole, the ground and the rope make a right-angled triangle, with the right angle at the foot of the pole. The pole is opposite the angle θ, so the reciprocal ratios give the two unknown sides directly: the cosecant for the rope and the cotangent for the distance along the ground.

1. The pole is opposite the angle θ and the rope is the hypotenuse, so sinθ = 1.5L for a rope of length L. Then L = 1.5sinθ = 1.5cosecθ.
2. The distance d from the foot of the pole to the peg is adjacent to θ, so tanθ = 1.5d, and d = 1.5tanθ = 1.5cotθ.
3. (a) At θ = π3, cosecπ3 = 2√3 and cotπ3 = 1√3. So L = 3√3 = √3, which is 1.73 m, and d = 1.5√3 = √32, which is 0.87 m.
4. With a 3 m rope, 1.5cosecθ = 3, so cosecθ = 2 and sinθ = 12. The angle is inside a right-angled triangle, so it is acute, and θ = π6.
5. (b) The rope meets the ground at π6, which is 30^°, and its peg is 1.5cotπ6 = 1.5√3 m from the pole, which is 2.60 m. Check: 1.5² + (1.5√3)² = 2.25 + 6.75 = 9 = 3².

answer(a) the rope is √3 m, which is 1.73 m, and the peg is √32 m, which is 0.87 m, from the pole; (b) π6, which is 30^°, with the peg 1.5√3 m, which is 2.60 m, from the pole

techniqueSecant, Cosecant and Cotangent · Exact Trigonometric Values in Radians

examsO-Level

The track, its horizontal distance and its rise make a right-angled triangle with θ at the bottom station. The horizontal distance is adjacent to θ and the track is the hypotenuse, so the track is the horizontal distance times secθ. The identity 1 + tan²θ = sec²θ turns the gradient into secθ directly.

1. The horizontal distance is adjacent to θ and the track is the hypotenuse, so cosθ = 240track, and the track is 240cosθ = 240secθ m.
2. The identity 1 + tan²θ = sec²θ gives sec²θ = 1 + (34)² = 1 + 916 = 2516.
3. (a) The angle is acute, so secθ is positive: secθ = 54. The negative root −54 is rejected, because the cosine of an acute angle is positive.
4. (b) The track is 240 × 54 = 300 m long, and it climbs 240tanθ = 240 × 34 = 180 m. Check: 240² + 180² = 57600 + 32400 = 90000 = 300².

answer(a) secθ = 54; (b) 300 m of track, climbing 180 m

techniqueThe Identity 1 + tan²θ = sec²θ · Secant, Cosecant and Cotangent

examsO-Level

To prove an identity, start from one side and reach the other with identities you already know, never working on both sides at once. The handbook's side is the more complicated, so start there: write the secant in terms of the cosine, combine it into one fraction, and use sin²θ + cos²θ = 1.

1. Work on the handbook's side only. Write the secant as one over the cosine and put the two terms over a common denominator: secθ − cosθ = 1cosθ − cosθ = 1 − cos²θcosθ.
2. The identity sin²θ + cos²θ = 1 gives 1 − cos²θ = sin²θ, so secθ − cosθ = sin²θcosθ.
3. (a) Divide by sinθ: sin²θcosθsinθ = sinθcosθ = tanθ. So the handbook's formula is F = Wtanθ, as required.
4. (b) At θ = π3, tanπ3 = √3, so F = 60√3 N, which is 103.9 N to 1 decimal place. Check with the forces: the tension in the cable is Wsecπ3 = 120 N, and its horizontal part, 120sinπ3 = 60√3 N, is what the rope must balance.

answer(a) secθ − cosθsinθ = sin²θcosθsinθ = tanθ, so F = Wtanθ; (b) 60√3 N, which is 103.9 N

techniqueProving a Trigonometric Identity · Trigonometric Identities · Secant, Cosecant and Cotangent

examsO-Level