01 · Match the coefficients
For 5sin x + 12cos x = R sin(x + α), give R, cos α and sin α.
Hint
Square and add the coefficients.
Worked solution
R = 13, cos α = 5/13 and sin α = 12/13. The phase is acute: α = arctan(12/5).
Understand · explore · practise
R formula for trigonometric expressions
Write a sin x + b cos x as one sine or cosine using the R formula. Match coefficients, choose the correct phase quadrant and understand amplitude and graph shifts.
Before you startCompound angle formulae, quadrant signs and graph translations
01 / Combine two waves of the same frequency
Their sum is one shifted wave.
The R formula rewrites a sine term and a cosine term with the same input angle as a single shifted sine or cosine. This makes the amplitude and phase easier to see and prepares an expression for equations or modelling.
a sin x + b cos x = R sin(x + α)
Choose R ≥ 0. When the expression is not identically zero, R is positive and α is fixed up to complete turns. Try both positive and negative coefficients in the model.
Add two waves point by pointExplore
Blue: a sin x. Green: b cos x. Gold: their sum. Same inputs, added heights.
R = 5; cos α = 3/5; sin α = 4/5.
Combined form: 5 sin(x + α), with α ≈ 53.1301°.
At x = 30°: 1.5000 + 3.4641 = 4.9641.
Decimal phase and point readouts are rounded. The coefficient equations specify the phase exactly. An amplitude bound need not be reached on a restricted interval.
02 / Match coefficients after expanding
Each original term determines one phase ratio.
Expand the proposed single sine:
R sin(x + α) = R cos α sin x + R sin α cos x
For equality at every x, match the coefficients of sin x and cos x:
R cos α = a
R sin α = b
Square and add. Since cos² α + sin² α = 1, we get R² = a² + b². Thus:
R = √(a² + b²)
cos α = a/R, sin α = b/R
The phase ratios require R > 0.
The two signed ratios determine the quadrant; a tangent ratio on its own does not.
Watch the component heights add to the combined wave
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Find amplitude and phase for positive coefficients
An acute phase is valid here because both ratios are positive.
Write 3sin x + 4cos x in the form R sin(x + α), 0° ≤ α < 360°.Worked example
R = √(3² + 4²) = 5
Use the positive square root.
cos α = 3/5 and sin α = 4/5
Both are positive, so α is in quadrant one.
α = arctan(4/3)
This exact inverse-trig expression selects the acute phase.
3sin x + 4cos x = 5sin(x + arctan(4/3))
Use consistent angle units when evaluating it.
The amplitude is 5, not 3 + 4 = 7. The sine and cosine components do not reach their maxima at the same input.
04 / Use coefficient signs to choose the phase quadrant
Do not accept the calculator branch without checking both ratios.
Write −3sin x + 4cos x as R sin(x + α), 0° ≤ α < 360°.Worked example
R = 5; cos α = −3/5 and sin α = 4/5
Cosine is negative and sine positive.
Let β = arctan(4/3), with β acute
This is the reference angle.
α = 180° − β
The required phase lies in quadrant two.
−3sin x + 4cos x = 5sin(x + 180° − β)
Expanding checks both coefficient signs.
For 3sin x − 4cos x the same reference angle gives the signed phase −β, or 360° − β in the interval [0°, 360°). For −3sin x − 4cos x it gives 180° + β.
05 / Use a minus-phase form when it is clearer
The sign inside the bracket changes a coefficient.
Write 12sin x − 5cos x as R sin(x − β), with β acute.Worked example
R sin(x − β) = R cos β sin x − R sin β cos x
Expand before matching.
R = 13; cos β = 12/13 and sin β = 5/13
Both ratios for β are positive.
β = arctan(5/12)
The phase in the plus form would be −β.
12sin x − 5cos x = 13sin(x − β)
Do not replace the minus by plus while keeping the same acute β.
06 / Choose a cosine form by matching again
The coefficient correspondence changes.
a sin x + b cos x = R cos(x − β)
R cos β = b, R sin β = a
Expand R cos(x − β) = R cos β cos x + R sin β sin x. Notice that the sine coefficient now matches R sin β.
Write 3cos x − √3 sin x as R cos(x + β), with β acute.Worked example
R = √(9 + 3) = 2√3
The amplitude is independent of the chosen sine/cosine form.
R cos β = 3 and R sin β = √3
The plus-angle cosine expansion supplies the minus sine term.
cos β = √3/2 and sin β = 1/2
Therefore β = π/6.
3cos x − √3 sin x = 2√3 cos(x + π/6)
Angles here are in radians.
07 / Handle zero coefficients and zero amplitude
Axis cases do not need division by a coefficient.
If one coefficient is zero, use the two matching equations directly. Do not divide b by a when a = 0.
4cos x = 4sin(x + 90°)
−4cos x = 4sin(x + 270°)
−3sin x = 3sin(x + 180°)
If a = b = 0, then R = 0 and the original expression is zero for every input. Any phase produces the same zero expression, so no phase is determined. The ratios a/R and b/R are undefined in this case.
08 / Read the graph shift from the bracket
A positive phase inside sine shifts the graph left.
For R sin(x + α), the unshifted sine graph is moved α units to the left, in the same angle units as x. Equivalent phases differing by a full turn describe the same curve.
Interpret 3sin 2x + 4cos 2x, with x in radians.Worked example
Let α = arctan(4/3)
The coefficient matching gives the same amplitude and phase as before.
3sin 2x + 4cos 2x = 5sin(2x + α)
Keep the original input 2x.
= 5sin(2(x + α/2))
Factor the coefficient of x before reading the shift.
Amplitude 5; period π; shift α/2 to the left
The shift in x is half the phase inside the angle.
Adding a constant afterwards moves the midline without changing the amplitude. Restricted-window maxima and minima require a separate interval check.
09 / Check the form by expanding or using diagnostic inputs
A plausible amplitude does not guarantee the right phase.
After finding a form, expand it back. Also check x = 0 and x = 90°: the original expression gives b and a respectively. These two checks expose coefficient swaps and many sign mistakes.
Why 3sin x + 4cos x cannot equal 5sin(x + arctan(3/4))Worked example
At x = 0 the original gives 4
The cosine coefficient is visible directly.
The proposed form gives 5 × 3/5 = 3
The phase ratios have been swapped.
The correct acute tangent is 4/3
It follows from sin α = 4/5 and cos α = 3/5.
The standard R formula combines terms with the same frequency. It cannot in general turn sin x + cos 2x into R sin(x + α). A sum with different frequencies needs a different treatment.
10 / Your turn
Always state the two signed coefficient equations.
02 · Quadrant two
Write −5sin x + 12cos x in the same form with 0° ≤ α < 360°.
Hint
Cosine of the phase is negative; sine is positive.
Worked solution
R = 13 and α = 180° − arctan(12/5), using degree-mode inverse tangent. The coefficient equations are cos α = −5/13 and sin α = 12/13.
03 · Quadrant three
Find the phase for −5sin x − 12cos x, using positive R and 0° ≤ α < 360°.
Hint
Both phase ratios are negative.
Worked solution
R = 13 and α = 180° + arctan(12/5).
04 · Quadrant four
Write 5sin x − 12cos x as R sin(x − β), with β acute.
Hint
Use the minus expansion.
Worked solution
13sin(x − β), where β = arctan(12/5), cos β = 5/13 and sin β = 12/13.
05 · Exact phase
Write √3 sin x + cos x as one sine in radians.
Hint
The amplitude is 2.
Worked solution
2sin(x + π/6), since cos(π/6) = √3/2 and sin(π/6) = 1/2.
06 · Cosine plus form
Write 2cos x − 2sin x as R cos(x + β), with β acute.
Hint
Both matching magnitudes are 2.
Worked solution
R = 2√2 and β = π/4 in radians: 2√2 cos(x + π/4).
07 · Cosine minus form
Write 8sin x + 15cos x as R cos(x − β), with β acute.
Hint
Here R sin β = 8 and R cos β = 15.
Worked solution
R = 17, β = arctan(8/15). Thus 17cos(x − β).
08 · Axis phase
Write −6cos x as R sin(x + α), with R > 0 and 0° ≤ α < 360°.
Hint
The phase sine is −1 and its cosine is 0.
Worked solution
6sin(x + 270°). Equivalently, 6sin(x − 90°).
09 · Zero expression
What are the amplitude and phase of 0sin x + 0cos x?
Hint
The resulting curve is identically zero.
Worked solution
R = 0. The phase is not determined; any phase gives the same zero curve. Do not compute 0/0.
10 · Period and shift
Let α = arctan(12/5). Find the amplitude, period and horizontal shift of 5sin 3x + 12cos 3x for radian x.
Hint
Write 13sin(3(x + α/3)).
Worked solution
Amplitude 13, period 2π/3 and shift α/3 to the left.
11 · Diagnose a swap
Why does 5sin x + 12cos x ≠ 13sin(x + arctan(5/12))?
Hint
Check x = 0.
Worked solution
The original is 12 at x = 0, but the proposed form is 5. The correct acute phase is arctan(12/5).
12 · Equivalent forms
Show that 4sin(x + π/3) and 4cos(x − π/6) are identical.
Hint
Use sin u = cos(π/2 − u), then cosine symmetry; or expand both.
Worked solution
Both expand to 2sin x + 2√3 cos x. The phase descriptions differ but the graph is the same.
13 · Different frequencies
Can sin x + cos 2x equal R sin(x + α) for all x? Give a contradiction.
Hint
Compare x = 0 and x = π.
Worked solution
The original equals 1 at both inputs. Any single sine of frequency 1 changes sign between inputs π apart, so it cannot equal 1 at both. No such constants R, α exist.
14 · Range versus a short interval
Does amplitude 5 alone prove that 3sin x + 4cos x reaches 5 on 0° ≤ x ≤ 10°?
Hint
The maximum occurs when x + α = 90° modulo 360°, where α = arctan(4/3).
Worked solution
No. The first maximum is at x = 90° − α ≈ 36.87°, outside the interval. The global amplitude bound does not certify attainment on this short interval.
11 / Recap
Amplitude from squares, phase from signed matching.
- Expand the requested sine or cosine form before matching coefficients.
- Use R = √(a² + b²) with R ≥ 0.
- Determine the phase quadrant from both signed ratios.
- Handle zero amplitude and axis phases separately.
- Factor any input multiplier before reading the horizontal shift.
- Check the result by expanding it back.
Section 1 of 11 · Combine two waves of the same frequency