Solving a sin x + b cos x = c with the R Form

One wave to solve, and its crest priced at R.

One wave, then a simple equation

Solve 3 sin x + 4 cos x = 2.5 for 0° ≤ x ≤ 360°. The left side mixes a sine and a cosine, so no single inverse function undoes it. Write it as one wave first.

From the harmonic form, 3 sin x + 4 cos x = 5 sin(x + α), with R = √(9 + 16) = 5 and tan α = 4/3, so α = 53.13° to 2 decimal places. The equation becomes 5 sin(x + 53.13°) = 2.5. Divide both sides by R = 5: sin(x + 53.13°) = 0.5.

That is an equation in one sine, of the kind already solved, except that the angle inside the sine is x + 53.13° rather than x.

The range moves with the bracket

Write u for the bracket: u = x + 53.13°. The equation is now sin u = 0.5. But the range was given for x, not for u.

Add 53.13° to every part of 0° ≤ x ≤ 360°: 53.13° ≤ u ≤ 413.13°. So the solutions for u must lie between 53.13° and 413.13°. They do not start at 0°.

Solve for u

sin u = 0.5 at u = 30° and at u = 180° − 30° = 150°, and again a full turn later, at 390° and 510°.

Now keep only the ones in the range 53.13° ≤ u ≤ 413.13°. 30° is below 53.13°, so it is dropped. 150° is in the range. 390° is in the range, because it is less than 413.13°. 510° is above 413.13°, so it is dropped. That leaves u = 150° and u = 390°.

−720°−360°0°360°720°sin x = 0.5: x = 30°, 150°n = 0n

sin x = 0.5 crosses the circle at 30° and 150°; turn n = 0 adds 360° × 0 to each, x = 30°, 150°

Set sin x = 0.5 and turn n to 1

The level line at a sine of 0.5 cuts the circle at two angles, 30° and 150°, and the wave on the right at those two angles in the first turn. Here the angle is u. Drag n to 1: both solutions move on a full turn, to 390° and 510°. Only 150° and 390° lie between 53.13° and 413.13°.

Subtract α

Each value of u is x + 53.13°, so x = u − 53.13°. From u = 150°, x = 150° − 53.13° = 96.87°. From u = 390°, x = 390° − 53.13° = 336.87°. To 1 decimal place the answers are 96.9° and 336.9°, and both are between 0° and 360°, as they must be.

Check by substitution in the original equation. At x = 96.87°, 3 sin 96.87° + 4 cos 96.87° = 2.5000, to 4 decimal places. At x = 336.87°, 3 sin 336.87° + 4 cos 336.87° = 2.5000 as well.

The dropped value u = 30° would give x = 30° − 53.13° = −23.13°. That is a solution of the equation, but it is outside the range asked for.

xy

The gold curve is y = 3 sin x + 4 cos x over one turn, with one square across for every 90°, and the plain straight line is y = 2.5. The wave has amplitude 5, and the line cuts it twice in the range, at the dots: x = 96.87° on the way down and x = 336.87° on the way up.

The greatest and least values

The same form gives the extremes. 5 sin(x + 53.13°) is greatest when the sine is 1, so the greatest value is R = 5, where x + 53.13° = 90°, at x = 36.87°. It is least when the sine is −1, so the least value is −5, where x + 53.13° = 270°, at x = 216.87°.

For 5 sin x + 12 cos x, R = √(25 + 144) = 13 and tan α = 12/5, so α = 67.38°. The least value is −13, where x + 67.38° = 270°, at x = 202.62°. Check: 5 sin 202.62° + 12 cos 202.62° = −13.

A constant added outside the wave lifts every value with it. 7 + 5 sin x + 12 cos x = 7 + 13 sin(x + 67.38°), so its greatest value is 7 + 13 = 20, at x = 22.62°, and its least is 7 − 13 = −6.

When the right side equals R, or is larger

Solve 8 sin x + 6 cos x = 10 for 0° ≤ x ≤ 360°. Here R = √(64 + 36) = 10 and tan α = 6/8, so α = 36.87°. The equation is 10 sin(x + 36.87°) = 10, so sin(x + 36.87°) = 1.

A sine equals 1 only at the top of the wave, once a turn: x + 36.87° = 90°, so x = 53.13°. The next one, 450°, gives x = 413.13°, outside the range. The line y = 10 touches the peak of the wave instead of cutting it, so there is one solution, not two. Check: 8 sin 53.13° + 6 cos 53.13° = 8 × 0.8 + 6 × 0.6 = 10.

If the right side is more than R, there is no solution at all. 3 sin x + 4 cos x = 6 would need sin(x + 53.13°) = 1.2, and no sine is more than 1.

The usual mistakes

Leaving the range at 0° to 360° for u. That keeps u = 30°, which gives x = −23.13°, outside the range, and loses u = 390°, which gives the answer 336.87°.

Stopping at the values of u. 150° and 390° are values of x + 53.13°; subtract 53.13° from each to get x.

Placing the greatest value at x = α. The peak is where x + α = 90°, at 36.87°, not at 53.13°.

Adding the coefficients for the greatest value. For 3 sin x + 4 cos x it is 5, not 7: the two waves never peak at the same x.

A buoy on two sets of waves

In the application below, the height of a buoy is √3 sin x + cos x meters, with x in seconds. It is 2 sin(x + π/6), so the greatest height is 2 m, and the time it is 1 m below the calm level comes from sin(x + π/6) = −½, with the bracket starting at π/6, not at 0.

Worked example: A Buoy Riding Two Sets of Waves at Once

Question A buoy rises and falls on two sets of waves at once. Its height above the calm-water level, x seconds after a stopwatch is started, is h = √3sin x + cos x meters. (a) Write h in the form Rsin(x + α), where R > 0 and 0 < α < π2. Find the greatest height of the buoy and the first time it reaches it. (b) When is the buoy first 1 m below the calm-water level? Give exact times, then times to 2 decimal places.

  1. 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.

    −2−101201234567seconds after the start, xheight (m), h√3 sin xcos xthe sumR cos α = √3, R sin α = 1: R = 2, α = pi/6
    −2−101201234567seconds after the start, xheight (m), h√3 sin xcos xthe sumR cos α = √3, R sin α = 1: R = 2, α = pi/6
    Rcosα = √3 and Rsinα = 1, so R = 2 and α = π6: the gold wave is the sum of the other two.

  2. 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.

    −2−101201234567seconds after the start, xheight (m), hRR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 m
    −2−101201234567seconds after the start, xheight (m), hRR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 m
    h = 2sin(x + π6), whose greatest value is R = 2 m.

  3. 3.(a) x = π2 − π6 = π3: the buoy is first 2 m up after π3 seconds, which is 1.05 seconds.

    −2−101201234567seconds after the start, xheight (m), hR1.05 sR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 s
    −2−101201234567seconds after the start, xheight (m), hR1.05 sR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 s
    (a) The peak comes when x + π6 = π2, at x = π3 ≈ 1.05 seconds.

  4. 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.

    −2−101201234567seconds after the start, xheight (m), hR1.05 s−1 mR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 ssin(x + pi/6) = −1/2, so x + pi/6 = 7 pi/6
    −2−101201234567seconds after the start, xheight (m), hR1.05 s−1 mR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 ssin(x + pi/6) = −1/2, so x + pi/6 = 7 pi/6
    Set h = −1: sin(x + π6) = −12, and the first angle after π6 is 7π6.

  5. 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.

    −2−101201234567seconds after the start, xheight (m), hR1.05 s−1 m3.14 sR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 ssin(x + pi/6) = −1/2, so x + pi/6 = 7 pi/6x = pi = 3.14 s
    −2−101201234567seconds after the start, xheight (m), hR1.05 s−1 m3.14 sR cos α = √3, R sin α = 1: R = 2, α = pi/6h = 2 sin(x + pi/6): greatest 2 mx + pi/6 = pi/2, so x = pi/3 = 1.05 ssin(x + pi/6) = −1/2, so x + pi/6 = 7 pi/6x = pi = 3.14 s
    (b) x = π ≈ 3.14 seconds, where the wave first meets the line h = −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

Common mistakes

  • Taking −π6 as the angle with a sine of −12, which gives x = −π3. That is before the stopwatch started; the angle x + π6 is at least π6, so the first solution in range is 7π6.
  • Giving π2 seconds for the greatest height, which is when sin x alone peaks. The combined wave is shifted by α = π6, so it peaks π6 sooner, at π3.

More radians and trigonometric identities problems, worked step by step →

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