Compound-Angle Formulae

Set A and B in degrees, or the way an exam sets them, by one function's value and a quadrant, and choose A + B or A − B. The picture reaches P in two walks, cos B along the turned axis and ±sin B across it; the table works the three formulae, and works the sine and cosine exactly in surds when A and B are whole multiples of 15°, the tangent too when the tangents involved exist.

Sine	$sin (A \pm B) = sin A cos B \pm cos A sin B$
Cosine, opposite sign	$cos (A \pm B) = cos A cos B \mp sin A sin B$
Tangent	$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$

Worked example — Compound-Angle Formulae

Compound angle	75°
Sine of A	0.7071
Cosine of A	0.7071
Sine of B	0.5
Cosine of B	0.866
Sine formula	0.9659
Cosine formula	0.2588
Tangent formula	3.7321

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Compound-Angle Formulae

The compound-angle formulae give the sine, cosine and tangent of A + B and of A − B from the sines and cosines of A and B.

Where the formula comes from

Turn the axes through A. The turned x-axis points along (cos A, sin A) and the turned y-axis along (−sin A, cos A). Measured in the turned frame P is at angle B, so it lies cos B along the first and sin B along the second.
$P = cos B (cos A, sin A) + sin B (- sin A, cos A)$
Read off P's coordinates. From the real x-axis P is at angle A + B, so its coordinates are cos(A + B) and sin(A + B). Each is two products: the orange legs across, the blue legs up.
$cos (A + B) sin (A + B) = cos A cos B - sin A sin B = sin A cos B + cos A sin B$
Replace B by −B. cos(−B) = cos B and sin(−B) = −sin B, so only the terms carrying sin B change sign.
$cos (A - B) sin (A - B) = cos A cos B + sin A sin B = sin A cos B - cos A sin B$
Divide the sine by the cosine. Dividing every term by cos A cos B turns the quotients into tangents and leaves a 1, which needs cos A and cos B to be non-zero.
$tan (A + B) = \frac{sin A cos B + cos A sin B}{cos A cos B - sin A sin B} = \frac{tan A + tan B}{1 - tan A tan B}$

Problems solved in full

An exact value from a sum of special angles 2 steps
Find the exact value of sin 75°.

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1. 75° = 45° + 30°, and both have exact sines and cosines. Expand:
  
  $sin 7 5^{\circ} = sin 4 5^{\circ} cos 3 0^{\circ} + cos 4 5^{\circ} sin 3 0^{\circ}$
2. Substitute, and put both terms over 4:
  
  $= \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} = \frac{6}{4} + \frac{2}{4}$
Answer

sin 75° = (√6 + √2)/4 ≈ 0.9659.

$sin 7 5^{\circ} = \frac{6 + 2}{4} \approx 0.9659$
An exact tangent from a difference 2 steps
Find tan 15° exactly, in the form a + b√3.

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1. 15° = 60° − 45°, with tan 60° = √3 and tan 45° = 1:
  
  $tan 1 5^{\circ} = \frac{tan 6 0 ^{\circ} - tan 4 5 ^{\circ}}{1 + tan 6 0 ^{\circ} tan 4 5 ^{\circ}} = \frac{3 - 1}{1 + 3}$
2. Rationalise by multiplying top and bottom by √3 − 1:
  
  $= \frac{( 3 - 1 ) ^{2}}{( 3 + 1 ) ( 3 - 1 )} = \frac{4 - 2 3}{2} = 2 - 3$
Answer

tan 15° = 2 − √3 ≈ 0.2679.

$tan 1 5^{\circ} = 2 - 3 \approx 0.2679$
A sine and a cosine in two quadrants 3 steps
sin A = 4/5 with A obtuse, and cos B = −5/13 with 180° < B < 270°. Find sin(A + B), cos(A + B) and tan(A + B), and the quadrant of A + B.

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1. First the two missing values, each from sin² + cos² = 1 with its quadrant's sign: cosine is negative in quadrant II and sine is negative in quadrant III.
  
  $cos A = - 1 - \frac{16}{25} = - \frac{3}{5}, sin B = - 1 - \frac{25}{169} = - \frac{12}{13}$
2. Then the sum formulae:
  
  $sin (A + B) cos (A + B) = \frac{4}{5} (- \frac{5}{13}) + (- \frac{3}{5}) (- \frac{12}{13}) = \frac{- 20 + 36}{65} = \frac{16}{65} = (- \frac{3}{5}) (- \frac{5}{13}) - \frac{4}{5} (- \frac{12}{13}) = \frac{15 + 48}{65} = \frac{63}{65}$
3. Both are positive. A + B lies between 270° and 450°, and with a positive sine and cosine it must be between 360° and 450°, which is quadrant I. Its tangent is the quotient:
  
  $tan (A + B) = \frac{16/65}{63/65} = \frac{16}{63}$
Answer

sin(A + B) = 16/65 ≈ 0.2462, cos(A + B) = 63/65 ≈ 0.9692, tan(A + B) = 16/63 ≈ 0.254, and A + B is in quadrant I.

$sin (A + B) = \frac{16}{65}, cos (A + B) = \frac{63}{65}, tan (A + B) = \frac{16}{63}$
An angle sum from two tangents 2 steps
tan A = 1/2 and tan B = 1/3, with A and B acute. Find A + B without a calculator.

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1. Put the two tangents into the formula:
  
  $tan (A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} = \frac{5/6}{5/6} = 1$
2. Both tangents are below 1, so A and B are each under 45° and A + B is between 0° and 90°. The only angle there with tangent 1 is 45°.
  
  $0^{\circ} < A + B < 9 0^{\circ}, tan (A + B) = 1 \Rightarrow A + B = 4 5^{\circ}$
Answer

A + B = 45°, though A ≈ 26.5651° and B ≈ 18.4349° are not special angles.

$A + B = 4 5^{\circ}$
When tan(A + B) has no value 3 steps
tan A = 2 and tan B = 1/2, with A and B acute. What happens to tan(A + B), and what is A + B?

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1. The formula's denominator is zero:
  
  $1 - tan A tan B = 1 - 2 \cdot \frac{1}{2} = 0$
2. That denominator is cos(A + B)/(cos A cos B), and cos A cos B is not zero, so
  
  $1 - tan A tan B = \frac{cos ( A + B )}{cos A cos B} = 0 \Rightarrow cos (A + B) = 0$
3. A + B is between 0° and 180°, and the only angle there with cosine 0 is 90°:
  
  $0^{\circ} < A + B < 18 0^{\circ} \Rightarrow A + B = 9 0^{\circ}$
Answer

tan(A + B) is undefined, because A + B = 90°. For acute A and B, tan A tan B = 1 always means the two angles add to 90°.

$A + B = 9 0^{\circ}, cos (A + B) = 0$
Back into one compound angle 2 steps
Find the exact value of cos 80° cos 20° + sin 80° sin 20°.

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1. cos · cos + sin · sin is the expansion of cos(A − B), here with A = 80° and B = 20°:
  
  $cos 8 0^{\circ} cos 2 0^{\circ} + sin 8 0^{\circ} sin 2 0^{\circ} = cos (8 0^{\circ} - 2 0^{\circ})$
2. and 80° − 20° = 60°:
  
  $= cos 6 0^{\circ} = \frac{1}{2}$
Answer

cos 80° cos 20° + sin 80° sin 20° = cos 60° = 1/2.

$cos 8 0^{\circ} cos 2 0^{\circ} + sin 8 0^{\circ} sin 2 0^{\circ} = \frac{1}{2}$
Proving an identity 3 steps
Prove that sin(A + B) + sin(A − B) = 2 sin A cos B, and check it at A = 45°, B = 30°.

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1. Expand both sines:
  
  $sin (A + B) + sin (A - B) = (sin A cos B + cos A sin B) + (sin A cos B - cos A sin B)$
2. The cos A sin B terms cancel:
  
  $= 2 sin A cos B$
3. At 45° and 30° the left side is sin 75° + sin 15°, and the right side is 2 sin 45° cos 30°:
  
  $\frac{6 + 2}{4} + \frac{6 - 2}{4} = \frac{6}{2} = 2 \cdot \frac{2}{2} \cdot \frac{3}{2}$
Answer

sin(A + B) + sin(A − B) = 2 sin A cos B for every A and B; at 45° and 30° both sides are √6/2 ≈ 1.2247.

$sin (A + B) + sin (A - B) = 2 sin A cos B$
The double angle falls out 2 steps
Put B = A in the sum formulae to find sin 2A and cos 2A, then evaluate both when sin A = 3/5 and A is obtuse.

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1. With B = A the two products in the sine are equal, and the cosine becomes a difference of squares:
  
  $sin 2 A = 2 sin A cos A, cos 2 A = cos^{2} A - sin^{2} A$
2. A is obtuse, so cos A = −4/5:
  
  $sin 2 A = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}, cos 2 A = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$
Answer

sin 2A = −24/25 = −0.96 and cos 2A = 7/25 = 0.28. 2A ≈ 286.2602° is in quadrant IV, where sine is negative and cosine positive.

$sin 2 A = - \frac{24}{25}, cos 2 A = \frac{7}{25}$
An equation with a compound angle 3 steps
Solve sin(x + 30°) = 2 cos x for 0° ≤ x < 360°.

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1. Expand the left side:
  
  $sin x cos 3 0^{\circ} + cos x sin 3 0^{\circ} = \frac{3}{2} sin x + \frac{1}{2} cos x$
2. Gather the cos x terms on the right and divide by cos x. It cannot be 0: then the equation would say sin x = 0 as well, and no angle has both.
  
  $\frac{3}{2} sin x = \frac{3}{2} cos x \Rightarrow tan x = 3$
3. Tangent repeats every 180°:
  
  $x = 6 0^{\circ}, x = 6 0^{\circ} + 18 0^{\circ} = 24 0^{\circ}$
Answer

x = 60° or x = 240°. At 60° both sides are 1, and at 240° both are −1.

$x = 6 0^{\circ}, 24 0^{\circ}$

Supplementary problems, answers only

Find the exact value of cos 105°.

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(√2 − √6)/4 ≈ −0.2588

$cos 10 5^{\circ} = \frac{2 - 6}{4}$
Find the exact value of tan 105°.

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−2 − √3 ≈ −3.7321

$tan 10 5^{\circ} = - 2 - 3$
sin A = 5/13 and cos B = 3/5, with A and B acute. Find cos(A − B).

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56/65 ≈ 0.8615

$cos (A - B) = \frac{56}{65}$
Write sin 3x cos x − cos 3x sin x as a single trigonometric function.

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sin 2x

$sin 3 x cos x - cos 3 x sin x = sin 2 x$
tan A = 3 and tan B = 1/2, with A and B acute. Find A − B.

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A − B = 45°

$A - B = 4 5^{\circ}$
Solve cos(x − 60°) = sin x for 0° ≤ x < 360°.

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x = 75° or x = 255°

$x = 7 5^{\circ}, 25 5^{\circ}$

The formulae are a rotation, written out 🖖

Measured from the axis turned through A, the point P sits at angle B, so it is reached by walking cos B along that axis to Q and then sin B at right angles. Each walk splits into an across part and an up part. Across: cos A cos B, then sin A sin B back the other way. Up: sin A cos B, then cos A sin B. P's across is cos(A + B) and its up is sin(A + B). Nothing in it needs an acute angle, because the products carry their own signs: when a leg turns round, its product changes sign. Press sin A = 4/5, cos B = −5/13: A is obtuse, B is in quadrant III, Q lands on the turned axis behind O, and the same sums still land on P at (0.9692, 0.2462). In matrix language, (cos(A + B), sin(A + B)) is the rotation matrix of A applied to (cos B, sin B), so the two formulae and “a turn by A, then by B, is a turn by A + B” are one statement. Ptolemy, in the second century, worked with chords instead of sines; his rules for the chord of a sum and of a difference of arcs correspond to the two sine formulae, and he used them to build a table of chords at every half degree.

sin(A + B) is almost never sin A + sin B 🖖

Press sin 75° = sin(45° + 30°). sin 45° + sin 30° = 0.7071 + 0.5 = 1.2071, more than any sine can be, while sin 75° = 0.9659. The formula weights each sine by the other angle's cosine, and a cosine is at most 1 in size. The two agree only when A, B or A + B is a whole number of turns, because their difference factors exactly: sin A + sin B − sin(A + B) = 4 sin(A/2) sin(B/2) sin((A + B)/2). So the guess fails at almost every pair; at 120° and 240° it holds, because A + B = 360° and both sides are 0.

The tangent formula can fail where the tangent exists 🖖

Press A = 90°, B = 45°. tan 135° = −1, but the formula asks for tan 90°, which has no value, so the formula has none either. Nothing is wrong with the angle; the formula divided by cos A cos B on the way, and cos 90° = 0. Divide top and bottom by tan A wherever tan A exists and is not 0 and the formula becomes (1 + cot A tan B)/(cot A − tan B), a form that also has a value at 90°: cot 90° = 0, so it reads 1/(0 − 1) = −1. The other failure is real. Press tan A = 2, tan B = 1/2: the denominator 1 − tan A tan B is 0. That denominator equals cos(A + B)/(cos A cos B), so it vanishes exactly when cos(A + B) does, and here A + B = 90°, where tangent is undefined. For acute angles, tan A tan B = 1 means B = 90° − A.

Learning path

Trigonometric identities and equations

References (2)

The sum and difference formulae for sine, cosine and tangent, the exact value of cos 75° from 45° + 30°, and the unit-circle distance proof of cos(α − β): J. Abramson, Precalculus 2e, OpenStax (2021), section 7.2, “Sum and Difference Identities”.
Insight block 1: Ptolemy (about AD 85 to 165) “used formulae for the Crd function which are analogous to our formulae for sin(a+b) and sin(a−b)” to build a table of chords “at intervals of 1/2 a degree”, in the first two books of the Almagest: J. J. O'Connor and E. F. Robertson, “Claudius Ptolemy”, MacTutor History of Mathematics Archive, University of St Andrews (April 1999).

By the 42Math team · Reviewed against published sources. Corrections welcome.