MathsTrigonometry › Compound angles and the harmonic form

Compound angles and the harmonic form

Sine refuses to distribute over addition, and the compound angle formulae are the machinery it demands instead. Out of them fall the double angle identities, a factory for new exact values, the half-angle rearrangements integration depends on, and the harmonic form, which folds any mix of sine and cosine into one wave whose size and timing you can read at sight.

Builds on Reciprocal and inverse trigonometric functions.

IN THIS TOPIC

  • Use the compound angle formulae, forwards and backwards.
  • Derive the double angle formulae by setting B = A, and choose between the three forms of cos 2A.
  • Rearrange cos 2A into the half-angle forms that integration later depends on.
  • Construct identity proofs of the kind the specification names.
  • Write a sin θ + b cos θ in harmonic form, and use it for equations and extremes.

COMMON MISCONCEPTION

sin (A + B) = sin A + sin B.

Angles that add

The first numbers anyone tries settle it. sin 90° is 1, while sin 30° + sin 60° is 1.37. Sine does not distribute over addition, and the booklet supplies the machinery that actually does the job,

sin(A±B)=sinAcosB±cosAsinB\sin (A ± B) = \sin A \cos B ± \cos A \sin BIN THE FORMULAE BOOKLET
cos(A±B)=cosAcosBsinAsinB\cos (A ± B) = \cos A \cos B ∓ \sin A \sin BIN THE FORMULAE BOOKLET
tan(A±B)=tanA±tanB1tanAtanB\tan (A ± B) = \frac{\tan A ± \tan B}{1 ∓ \tan A \tan B}IN THE FORMULAE BOOKLET

Cosine's middle sign flips. That single detail is the one most often mis-copied under time pressure, and it turns an otherwise perfect page into nothing. All three have geometrical proofs built from a pair of right-angled triangles, and the specification expects you to know they are provable even though the booklet hands them over.

Sine of x plus 60 degrees against sine x plus sine of 60 degrees: a shifted wave within plus and minus 1 against a raised wave climbing to 1.87sin x + sin 60°: escapes ±1sin (x + 60°): a shifted wavenot the same function, not even close
FIG. 1The two candidates, plotted. sin (x + 60°) is a shifted wave inside ±1; sin x + sin 60° rides up past 1.8. Nothing about them matches.

WORKED EXAMPLE

A new exact value

Find the exact value of sin 75°.

Split 75° into angles you already know. sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°.

Substituting the exact values gives (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.

Double angles, and the half-angle trick

Set B = A and the compound formulae collapse into the double angle formulae,

sin2A=2sinAcosA\sin 2A = 2 \sin A \cos ANOT IN THE BOOKLET — LEARN IT
cos2A=cos2Asin2A=2cos2A1=12sin2A\cos 2A = \cos^{2}A − \sin^{2}A = 2\cos^{2}A − 1 = 1 − 2\sin^{2}ANOT IN THE BOOKLET — LEARN IT
tan2A=2tanA1tan2A\tan 2A = \frac{2\tan A}{1 − \tan^{2}A}NOT IN THE BOOKLET — LEARN IT

Here is the thing worth writing on the front of your revision card. The booklet gives you the compound formulae and does not give you these. Should one of them desert you in the exam, set B = A in the printed formula and derive it on the spot in one line.

Rearranging the last two forms of cos 2A is how half angles arrive. cos2A = (1 + cos 2A)/2 and sin2A = (1 − cos 2A)/2, which the integration unit then uses constantly. Choose whichever version of cos 2A leaves you holding the ratio the question already contains, and you save yourself three lines of algebra every time.

GUIDED PRACTICE

A proof from the specification's shelf

Prove that cos x cos 2x + sin x sin 2x ≡ cos x, before opening the working.

Show the working

The left-hand side is the expansion of cos (2x − x), the compound cosine formula read from right to left.

cos (2x − x) = cos x, and the identity is proved. ∎

Everything rested on spotting a formula running backwards. Expanding into single angles also works, in five lines instead of two.

The harmonic form

Any combination a sin θ + b cos θ is secretly one wave. Expand R sin (θ + α), match the coefficients of sin θ and cos θ, and out come R = √(a2 + b2) and tan α = b/a. That rewrite is the harmonic form, and it hands over the extremes and every solution of a cos θ + b sin θ = c in a single move.

The wave 3 sine theta plus 4 cos theta: a single sine wave of amplitude 5, inside its dashed envelope at plus and minus 5peak 5 at θ = 0.643 sin θ + 4 cos θ
FIG. 23 sin θ + 4 cos θ, drawn. One sinusoid of amplitude exactly 5, peaking at θ = 0.64, precisely where R sin (θ + α) says it should.

WORKED EXAMPLE

Fold, then read everything off

Express 3 sin θ + 4 cos θ in the form R sin (θ + α) with R > 0 and 0 < α < π/2, and state the maximum value and where it first occurs.

R = √(9 + 16) = 5, and tan α = 4/3 gives α = 0.927.

So 3 sin θ + 4 cos θ = 5 sin (θ + 0.927).

The maximum is 5, reached where the sine hits 1. That needs θ + 0.927 = π/2, so θ = 0.644.

No calculus was needed for that maximum, and none ever is once the expression has become a single wave.

INDEPENDENT PRACTICE

An equation through the fold

Using the form above, solve 3 sin θ + 4 cos θ = 2.5 for 0 ≤ θ < 2π.

Show the working

The equation becomes 5 sin (θ + 0.927) = 2.5, so sin (θ + 0.927) = ½.

With φ = θ + 0.927 running over (0.927, 0.927 + 2π), the solutions of sin φ = ½ in range are φ = 5π/6 and π/6 + 2π.

Translating back gives θ = 1.69 and 5.88 (3 significant figures).

The widen, substitute and translate routine from the equations lesson ran unchanged. All the harmonic form did was make the equation solvable in the first place.

ASSESSMENT FOCUS

  • Copy the compound formulae out of the booklet carefully. Cosine's middle sign is the opposite of the one inside the bracket.
  • The booklet does not print the double angle formulae. If one has gone, set B = A in the printed compound formula and derive it.
  • Pick the cos 2A form that leaves you with the ratio the question already uses. The wrong choice adds three lines of pointless algebra.
  • In identity proofs work down one side only, name each formula as you use it, and finish with ≡.
  • For harmonic form, R = √(a² + b²) every time. α comes from tan α, checked against the signs of a and b before you commit to a quadrant.
  • Maximum and minimum of a sin θ + b cos θ are ±R. No differentiation is required, and stating that is the expected method.

CHECK YOURSELF

Express 5 cos θ − 12 sin θ in the form R cos (θ + α) with R > 0 and 0 < α < π/2, and write down the minimum value of the expression.

Show a hint

Expand R cos (θ + α) and match both coefficients.

Show the answer

Matching gives R cos α = 5 and R sin α = 12, so R = √(25 + 144) = 13 and tan α = 12/5, α = 1.176.

So 5 cos θ − 12 sin θ = 13 cos (θ + 1.176).

The minimum is −13, where the cosine reaches −1. The 5-12-13 triangle is what made R come out whole.

Angles add through the compound formulae, never through the functions themselves.

a sin θ + b cos θ is one wave of amplitude √(a² + b²); fold first, then read off extremes and roots.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the compound angles and the harmonic form questions page.

CHECK YOUR PROGRESS

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  • Use the compound angle formulae, forwards and backwards.
  • Derive the double angle formulae by setting B = A, and choose between the three forms of cos 2A.
  • Rearrange cos 2A into the half-angle forms that integration later depends on.
  • Construct identity proofs of the kind the specification names.
  • Write a sin θ + b cos θ in harmonic form, and use it for equations and extremes.

Open the full revision checklist to see every objective in the course in one place.