The compound angle formulae for sin(A ±B), cos(A ±B) and tan(A ±B), and a geometric proof

Exact values, and ratios of compound angles from given ratios and quadrants

The double-angle formulae for sin 2A, cos 2A (three forms) and tan 2A

Half-angle results such as cos² A = ½(1 + cos 2A)

Solving equations and proving identities with these formulae, in degrees or radians

Writing acos θ+ bsin θ as Rcos(θ±α) or Rsin(θ±α)

Solving acos θ+ bsin θ= c and finding greatest and least values, including in context

1. The compound angle formulae sin(A ±B) = sin Acos B ±cos Asin B and cos(A ±B) = cos Acos B ∓sin Asin B give exact values, turn into the double-angle formulae when B = A, and, read backwards, turn acos θ+ bsin θ into a single wave.

2. tan(A ±B) = (tan A ±tan B)/(1 ∓tan Atan B). All three compound angle formulae are in the formulae booklet.

3. Double angles (not in the booklet): sin 2A = 2sin Acos A; cos 2A = cos² A - sin² A = 2cos² A - 1 = 1 - 2sin² A; tan 2A = 2tan A/(1 - tan² A).

4. For equations, expand, then collect into a single ratio (often tan θ) or a quadratic in one ratio; choose the form of cos 2A that matches the rest of the equation.

5. R-form: acos θ+ bsin θ= Rcos(θ- α) with R = √(a² + b²) and tan α= b/a. Expand the target form and compare coefficients to get the signs right.

6. Rsin(…) and Rcos(…) lie between -R and R; the maximum R occurs when the bracket is 90° (for sin) or 0° (for cos).

It is tempting to think that sin(A + B) = sin A + sin B, but a single test destroys the idea: with A = B = 90°, the left side is sin 180° = 0 while the right side is 2. The sine of a sum mixes sines and cosines, and a picture shows exactly how.

In the figure, the angle A + B is built from two right-angled triangles. OP has length 1, so the height of P is sin(A + B) and its horizontal distance from O is cos(A + B). In triangle OQP, OQ = cos B and QP = sin B. Then QM = OQsin A = sin Acos B. Because QT is parallel to OM and QP is perpendicular to OQ, the angle TPQ is also A, so PT = QPcos A = cos Asin B. Adding the two heights gives sin(A + B) = sin Acos B + cos Asin B. The horizontal distances give ON = OM - TQ, which is cos(A + B) = cos Acos B - sin Asin B.

Replacing B by -B, with cos(-B) = cos B and sin(-B) = -sin B, gives the "minus" versions, and dividing the sine formula by the cosine formula (then dividing top and bottom by cos Acos B) gives the tan formula.

WORKED EXAMPLE · QUESTION

Without a calculator, find the exact values of (a) sin 105°; (b) tan 75°, giving your answer in the form p + q√3.

SOLUTION

(a) Split 105° into angles with known exact ratios: sin(60° + 45°) = sin 60°cos 45° + cos 60°sin 45° = √3/2·√2/2 + ½·√2/2 = (√6 + √2)/4.

(b) tan 75° = tan(45° + 30°) = (1 + 1/√3)/(1 - 1/√3) = (√3 + 1)/(√3 - 1) (multiplying top and bottom by √3). Rationalise: ((√3 + 1)²)/(3 - 1) = (4 + 2√3)/2 = 2 + √3. Check: 2 + √3 ≈3.7321, which matches tan 75° on a calculator.

A typical question gives one ratio of each angle and says which quadrant it lies in. To use the formulae you need both the sine and the cosine of each angle. Draw a right-angled triangle (or use sin² + cos² = 1) to find the size of the missing ratio, then use the quadrant to fix its sign. Acute angles have every ratio positive; obtuse angles (90° to 180°) have sin positive and cos and tan negative. Never find the angles on a calculator and round them: the question asks for exact values.

WORKED EXAMPLE · QUESTION

The angle A is acute with tan A = (8/15). The angle B is obtuse with sin B = (20/29). Find the exact values of (a) sin(A + B); (b) cos(A - B); (c) tan(A + B).

SOLUTION

Complete the ratios. For A: hypotenuse √(8² + 15²) = 17, so sin A = (8/17), cos A = (15/17). For B: third side √(29² - 20²) = 21, and cos is negative for an obtuse angle, so cos B = -(21/29), tan B = -(20/21).

(a) sin(A + B) = (8/17)·(-21)/29 + (15/17)·(20/29) = (-168 + 300)/493 = (132/493).

(b) cos(A - B) has a plus sign in the middle: (15/17)·(-21)/29 + (8/17)·(20/29) = (-315 + 160)/493 = -(155/493).

(c) tan(A + B) = ((8/15) - (20/21))/(1 + (8/15)·(20/21)) = (-(44/105))/(475/315) = -(132/475). Check: cos(A + B) = (-315 - 160)/493 = -(475/493), and sin÷cos= -(132/475). Since sin(A + B) > 0 and cos(A + B) < 0, the angle A + B is obtuse.

Expand every compound angle, using the exact values of the fixed angle (for example cos 30° = √3/2). Collect the sin θ terms on one side and the cos θ terms on the other, then divide by cos θ to reach tan θ= k. Dividing is allowed provided cos θ= 0 is not a solution, so check that by substituting. tan has period 180°, so in 0° ≤θ< 360° there are two solutions: the principal value and that value plus 180°.

WORKED EXAMPLE · QUESTION

Solve, for 0° ≤θ< 360°, the equation sin(θ+ 30°) = 2cos θ.

SOLUTION

Expand: sin θcos 30° + cos θsin 30° = √3/2sin θ+ ½cos θ. Multiply by 2 and collect: √3sin θ+ cos θ= 4cos θ⇒√3sin θ= 3cos θ. Divide by √3cos θ (cos θ= 0 would force sin θ= 0 too, which is impossible): tan θ= 3/√3 = √3. Principal value 60°, then add 180°: θ= 60° or 240°.

Put B = A in the addition formulae. sin(A + A) = sin Acos A + cos Asin A = 2sin Acos A, and cos(A + A) = cos² A - sin² A. Replacing sin² A by 1 - cos² A, or cos² A by 1 - sin² A, gives two more versions of cos 2A. The formulae work for any angle and its double: sin 6x = 2sin 3xcos 3x, and cos x = 1 - 2sin²x/2.

The rearranged forms are worth seeing as graphs. y = cos² x is a cosine wave of half the period, squeezed between 0 and 1 and centred on y = ½, which is exactly what ½(1 + cos 2x) describes.

WORKED EXAMPLE · QUESTION

Given that cos x = ⅓ and 270° < x < 360°, find the exact values of (a) sin 2x; (b) cos 2x; (c) tan 2x.

SOLUTION

sin x is negative in the fourth quadrant: sin x = -√(1 - (1/9)) = -2√2/3.

(a) sin 2x = 2(-2√2/3)(⅓) = -4√2/9.

(b) The cos-only form needs no square roots: cos 2x = 2·(1/9) - 1 = -(7/9).

(c) tan 2x = (-4√2/9)/(-(7/9)) = 4√2/7. Check with the tan formula: tan x = -2√2, so (2(-2√2))/(1 - 8) = 4√2/7. There is no need to find 2x or its quadrant first: the formulae carry the signs.

The aim is always one trigonometric function of one angle. If the equation also contains sin θ or cos θ, replace the double angle using the formulae, and choose the form that matches:

cos 2θ with sin θ elsewhere: use cos 2θ= 1 - 2sin²θ. With cos θ: use 2cos²θ- 1. The result is a quadratic.

sin 2θ with sin θ or cos θ: write sin 2θ= 2sin θcos θ and factorise. Never divide by sin θ or cos θ, which loses the solutions where it is zero.

If every term already has the same double angle (for example 3sin 2θ= cos 2θ), keep it: solve for 2θ over the doubled interval, then halve.

Reject any root of the quadratic outside -1 ≤sin θ≤1, and say why.

WORKED EXAMPLE · QUESTION

Solve, for 0° ≤θ< 360°, the equation 3cos 2θ= 5sin θ- 1.

SOLUTION

The other term is in sin θ, so use cos 2θ= 1 - 2sin²θ: 3 - 6sin²θ= 5sin θ- 1. Rearrange and factorise: 6sin²θ+ 5sin θ- 4 = 0 ⇒(2sin θ- 1)(3sin θ+ 4) = 0. So sin θ= ½, or sin θ= -(4/3), which has no solutions since -1 ≤sin θ≤1. Solving sin θ= ½: θ= 30° or 150°.

Start from the more complicated side and work towards the other. Do not treat the identity as an equation: never do the same thing to both sides. A double angle on one side and single angles on the other tells you which formula to use, and you should choose the form of cos 2A that cancels something: 1 + cos 2A = 2cos² A and 1 - cos 2A = 2sin² A are the most useful. Finish with "= RHS" or "≡tan x as required".

Putting A = θ/2 in the cosine formulae gives cos θ= 2cos²θ/2 - 1 = 1 - 2sin²θ/2, and so the half-angle results below.

WORKED EXAMPLE · QUESTION

(a) Prove that sin 2x/(1 + cos 2x) ≡tan x, for cos x ≠0. (b) Hence find the exact value of tan 22.5°, in the form p + q√2. (c) Prove that cos⁴ x - sin⁴ x ≡cos 2x.

SOLUTION

(a) Use the form of cos 2x that cancels the 1: LHS = 2sin xcos x/(1 + (2cos² x - 1)) = 2sin xcos x/(2cos² x) = sin x/cos x = tan x as required.

(b) Put x = 22.5°, so 2x = 45°: tan 22.5° = (√2/2)/(1 + √2/2) = √2/(2 + √2) = (√2(2 - √2))/(4 - 2) = √2 - 1 (≈ 0.4142, matching the calculator). "Hence" means the identity in (a) must be used.

(c) Difference of two squares, then sin² + cos² = 1: (cos² x + sin² x)(cos² x - sin² x) = 1 ×(cos² x - sin² x) = cos 2x.

Add a cosine wave and a sine wave of the same period and something remarkable happens: the result is a single wave of the same period, with a bigger amplitude R and a horizontal shift α. The figure shows 3cos θ+ 4sin θ, which is exactly 5cos(θ- 53.1°). Writing the sum this way is called the R-form (or harmonic form).

The method is always the same three lines:

expand the required form, for example Rcos(θ+ α) = Rcos θcos α- Rsin θsin α;

compare coefficients of cos θ and sin θ to get Rcos α= … and Rsin α= … (both positive when 0 < α< 90° and R > 0);

square and add for R = √(a² + b²); divide for tan α.

Use the form the question gives. As a guide, these choices keep α acute: acos+ bsin→Rcos(θ- α); acos- bsin→Rcos(θ+ α); bsin+ acos→Rsin(θ+ α); bsin- acos→Rsin(θ- α).

To solve acos θ+ bsin θ= c, write it as Rcos(θ- α) = c, so cos(θ- α) = c/R.

Shift the interval by α before finding the values of the bracket, then add α back (with a "+" form, subtract it). There are no solutions if |c| > R.

WORKED EXAMPLE · QUESTION

(a) Express 5cos θ- 2sin θ in the form Rcos(θ+ α), where R > 0 and 0 < α< 90°. Give the exact value of R and the value of α to 1 decimal place. (b) Hence solve, for 0 ≤θ< 360°, the equation 5cos θ- 2sin θ= 3, giving answers to 1 decimal place. (c) Write down the maximum value of 5cos θ- 2sin θ and the value of θ in this interval at which it occurs.

SOLUTION

(a) Rcos θcos α- Rsin θsin α, so Rcos α= 5 and Rsin α= 2. R = √(25 + 4) = √29; tan α= (2/5) ⇒ α= 21.8°.

(b) cos(θ+ 21.8°) = 3/√29, with 21.8° ≤θ+ 21.8° < 381.8°. The values of the bracket in this interval are 56.15° and 360° - 56.15° = 303.85°. Subtract the unrounded α: θ= 34.3° or 282.1°.

(c) The maximum of √29cos(…) is √29, when the bracket is 0° or 360°. Here θ+ 21.8° = 360°, so the maximum is √29 ≈5.39 at θ= 338.2°.

Since -1 ≤cos(…) ≤1, the expression Rcos(θ±α) lies between -R and R, and the same holds for Rsin(θ±α). The value of θ at the maximum comes from setting the bracket equal to 0° for cos (or 90° for sin); at the minimum, 180° for cos (or 270° for sin). Then solve for θ.

Linked expressions follow from this. For k + Rsin(…) the range is k - R to k + R. For a reciprocal such as m/(k + Rsin(…)) with k > R, the greatest value occurs when the denominator is least. For a squared term such as (Rsin(…))², the least value is 0, not -R². Contexts such as tides, temperatures and the heights of rotating objects use exactly this: the "greatest height" is k + R.

WORKED EXAMPLE · QUESTION

A rectangle PQRS has PQ = 8 cm and QR = 3 cm. It stands in a vertical plane with the corner P on a horizontal line and PQ at an angle θ to the line, where 0° < θ< 90°. The corner S is on the same side of PQ as R, above the line. (a) Show that the height of R above the line is h cm, where h = 8sin θ+ 3cos θ. (b) Express h in the form Rsin(θ+ α), where R > 0 and 0 < α< 90°. Give R exactly and α to 2 decimal places. (c) Find the greatest possible height of R above the line, and the value of θ at which it occurs. (d) Find the value of θ for which R is 7 cm above the line.

SOLUTION

(a) Q is 8sin θ above P. QR is perpendicular to PQ, so it makes angle θ with the vertical and rises 3cos θ. So h = 8sin θ+ 3cos θ as required.

(b) Rsin(θ+ α) = Rsin θcos α+ Rcos θsin α, so Rcos α= 8, Rsin α= 3. R = √73, α= 20.56° (tan α= (3/8)).

(c) Greatest when sin(θ+ α) = 1: h_(max) = √73 ≈ 8.54 cm when θ+ 20.56° = 90°, i.e. θ= 69.4°.

(d) sin(θ+ α) = 7/√73 ⇒θ+ α= 55.01° or 124.99°, so θ= 34.46° or 104.43°. Only θ= 34.5° lies in 0° < θ< 90°. The second value solves the equation but not the problem: at θ= 104.4° the side PQ would lean back past the vertical.

P3.3 Reciprocal and inverse trigonometric functions

Exact values, CAST and solving sin θ= k come from earlier units; sec, cosec, cot and sec²θ= 1 + tan²θ from P3.3 are often mixed into the same identity or equation as the formulae here.

P3.6 Differentiation and P4 integration

The derivatives of sin x and cos x rest on the compound angle formulae: a first-principles argument expands sin(x + h). In P4, sin² x and cos² x are integrated by first writing them as ½(1 ∓cos 2x).

Mechanics and Physics

Resolving forces gives terms like Pcos θ+ Qsin θ, and the R-form finds the angle that makes such a force greatest or least; projectile ranges use 2sin θcos θ= sin 2θ. Adding two oscillations of the same frequency gives a single oscillation: the R-form is the algebra of superposition, with R the resultant amplitude and α the phase.

1. Find the exact value of cos 15°.

2. Simplify 2sin 3xcos 3x.

3. Write cos² x in terms of cos 2x.

4. Find the greatest and least values of 1/(20 + 12sin θ- 5cos θ).

5. Solve sin 2θ= sin θ for 0° ≤θ< 360°.

6. Express 3cos θ+ 4sin θ in the form Rcos(θ- α) and state its minimum value.

1. cos(45° - 30°) = cos 45°cos 30° + sin 45°sin 30° = √2/2·√3/2 + √2/2·½ = (√6 + √2)/4.

2. It is sin 2A with A = 3x, so it equals sin 6x.

3. From cos 2x = 2cos² x - 1: cos² x = ½(1 + cos 2x).

4. 12sin θ- 5cos θ= 13sin(θ- α), with R = √(144 + 25) = 13. The denominator ranges from 20 - 13 = 7 to 20 + 13 = 33, so the greatest value is (1/7) and the least is (1/33).

5. 2sin θcos θ- sin θ= 0, so sin θ(2cos θ- 1) = 0. sin θ= 0 gives 0° and 180°; cos θ= ½ gives 60° and 300°. Four solutions: 0°, 60°, 180°, 300°.

6. Rcos α= 3, Rsin α= 4, so R = 5 and tan α= (4/3), α= 53.1°. The expression is 5cos(θ- 53.1°), with minimum -5 (when θ- 53.1° = 180°, i.e. θ= 233.1°).