Maths › Trigonometry › 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 correct machinery it demands instead. From them fall the double angle identities, a factory for new exact values, and the harmonic form, which folds any mix of sine and cosine into a single wave whose size and timing can be read at sight.

Year 12-13EDEXCEL 9MA0 5.6, 5.8

Builds on Reciprocal and inverse trigonometric functions.

IN THIS TOPIC

  • Use the compound and double angle formulae, forwards and backwards.
  • 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.

WHAT YOU PROBABLY THINK

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

Angles that add

The opening lie fails on the first number tried: sin 90° is 1, while sin 30° + sin 60° is 1.37. The truthful machinery, printed in the booklet, is the family of compound angle formulae,

sin (A ± B) = sin A cos B ± cos A sin BIN THE FORMULAE BOOKLET
cos (A ± B) = cos A cos B ∓ sin A sin BIN THE FORMULAE BOOKLET

with a tangent version alongside, and with cosine's middle sign flipped, the detail most often mis-copied.

The curves of sine of x plus 60 degrees and of sine x plus sine of 60 degrees: the first is a shifted sine wave staying within plus and minus 1, the second is a raised wave climbing to 1.87, so the two expressions are nothing like equalsin x + sin 60°: escapes ±1sin (x + 60°): a shifted wavenot the same function, not even close
FIG. 1The lie, 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 into known angles: sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°.

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

The exact-value table only ever had five entries; the compound formulae let those five breed every sum and difference of themselves.

Setting B = A produces the double angle formulae, on the must-learn list: sin 2A = 2 sin A cos A, and cos 2A in three interchangeable forms, cos2A − sin2A, 2cos2A − 1, and 1 − 2sin2A, each one substitution from the others via the Pythagorean identity.

YOUR TURN

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 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. ∎

Recognising a formula running backwards is the entire skill; expanding everything 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: expanding R sin (θ + α) and matching coefficients gives R = √(a2 + b2) and tan α = b/a. The rewrite is called the harmonic form, and it hands over the maximum, the minimum and every solution of a cos θ + b sin θ = c in one move.

The wave 3 sine theta plus 4 cos theta drawn with the dashed envelope at plus and minus 5: the sum of two waves is a single sine wave of amplitude 5, peaking at theta about 0.64peak 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, where the sine reaches 1: θ + 0.927 = π/2, so θ = 0.644.

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

TRY IT UNSEEN

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: θ = 1.69 and 5.88 (3 significant figures).

The widen-substitute-translate routine from the equations lesson runs unchanged; the harmonic form's only job was to make the equation solvable at all.

THE EXAM BIT

  • Quote compound formulae from the booklet carefully; cosine's middle sign is opposite to the one in the bracket.
  • Derive double angles by setting B = A, and pick the cos 2A form that suits the target expression.
  • In identity proofs work from one side, name each formula used, and finish with the identity symbol.
  • For harmonic form, R is √(a² + b²) always, and α comes from tan α with a quadrant check against the signs of a and b.
  • Max and min of a sin θ + b cos θ are ±R, no differentiation required, and saying so 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, and the 5-12-13 triangle made R exact.

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.

CHECK YOUR PROGRESS

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  • Use the compound and double angle formulae, forwards and backwards.
  • 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.

No animated video for this topic yet; these notes stand alone.