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.

Loading interactive simulation...

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.
$\begin{aligned}\cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \sin(A+B)&=\sin A\cos B+\cos A\sin B\end{aligned}$
Replace B by −B. cos(−B) = cos B and sin(−B) = −sin B, so only the terms carrying sin B change sign.
$\begin{aligned}\cos(A-B)&=\cos A\cos B+\sin A\sin B\\ \sin(A-B)&=\sin A\cos B-\cos A\sin B\end{aligned}$
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°.

Open this problem in the tool
1. 75° = 45° + 30°, and both have exact sines and cosines. Expand:
  
  $\sin75^{\circ}=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}$
2. Substitute, and put both terms over 4:
  
  $=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}$
Answer

sin 75° = (√6 + √2)/4 ≈ 0.9659.

$\sin75^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0{.}9659$
An exact tangent from a difference 2 steps
Find tan 15° exactly, in the form a + b√3.

Open this problem in the tool
Show solution
1. 15° = 60° − 45°, with tan 60° = √3 and tan 45° = 1:
  
  $\tan15^{\circ}=\frac{\tan60^{\circ}-\tan45^{\circ}}{1+\tan60^{\circ}\tan45^{\circ}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}$
2. Rationalise by multiplying top and bottom by √3 − 1:
  
  $=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
Answer

tan 15° = 2 − √3 ≈ 0.2679.

$\tan15^{\circ}=2-\sqrt{3}\approx0{.}2679$
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.

Open this problem in the tool
Show solution
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.
  
  $\cos A=-\sqrt{1-\tfrac{16}{25}}=-\tfrac{3}{5},\qquad\sin B=-\sqrt{1-\tfrac{25}{169}}=-\tfrac{12}{13}$
2. Then the sum formulae:
  
  $\begin{aligned}\sin(A+B)&=\tfrac{4}{5}\left(-\tfrac{5}{13}\right)+\left(-\tfrac{3}{5}\right)\left(-\tfrac{12}{13}\right)=\tfrac{-20+36}{65}=\tfrac{16}{65}\\ \cos(A+B)&=\left(-\tfrac{3}{5}\right)\left(-\tfrac{5}{13}\right)-\tfrac{4}{5}\left(-\tfrac{12}{13}\right)=\tfrac{15+48}{65}=\tfrac{63}{65}\end{aligned}$
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.

$\sin(A+B)=\tfrac{16}{65},\quad\cos(A+B)=\tfrac{63}{65},\quad\tan(A+B)=\tfrac{16}{63}$
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.

Open this problem in the tool
Show solution
1. Put the two tangents into the formula:
  
  $\tan(A+B)=\frac{\tfrac{1}{2}+\tfrac{1}{3}}{1-\tfrac{1}{2}\cdot\tfrac{1}{3}}=\frac{5/6}{5/6}=1$
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°.
  
  $0^{\circ}<A+B<90^{\circ},\quad\tan(A+B)=1\ \Rightarrow\ A+B=45^{\circ}$
Answer

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

$A+B=45^{\circ}$
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?

Open this problem in the tool
Show solution
1. The formula's denominator is zero:
  
  $1-\tan A\tan B=1-2\cdot\tfrac{1}{2}=0$
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°:
  
  $0^{\circ}<A+B<180^{\circ}\ \Rightarrow\ A+B=90^{\circ}$
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°.

$A+B=90^{\circ},\qquad\cos(A+B)=0$
Back into one compound angle 2 steps
Find the exact value of cos 80° cos 20° + sin 80° sin 20°.

Open this problem in the tool
Show solution
1. cos · cos + sin · sin is the expansion of cos(A − B), here with A = 80° and B = 20°:
  
  $\cos80^{\circ}\cos20^{\circ}+\sin80^{\circ}\sin20^{\circ}=\cos(80^{\circ}-20^{\circ})$
2. and 80° − 20° = 60°:
  
  $=\cos60^{\circ}=\tfrac{1}{2}$
Answer

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

$\cos80^{\circ}\cos20^{\circ}+\sin80^{\circ}\sin20^{\circ}=\tfrac{1}{2}$
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°.

Open this problem in the tool
Show solution
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°:
  
  $\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{6}}{2}=2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}$
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.

Open this problem in the tool
Show solution
1. With B = A the two products in the sine are equal, and the cosine becomes a difference of squares:
  
  $\sin2A=2\sin A\cos A,\qquad\cos2A=\cos^{2}A-\sin^{2}A$
2. A is obtuse, so cos A = −4/5:
  
  $\sin2A=2\cdot\tfrac{3}{5}\cdot\left(-\tfrac{4}{5}\right)=-\tfrac{24}{25},\qquad\cos2A=\tfrac{16}{25}-\tfrac{9}{25}=\tfrac{7}{25}$
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.

$\sin2A=-\tfrac{24}{25},\qquad\cos2A=\tfrac{7}{25}$
An equation with a compound angle 3 steps
Solve sin(x + 30°) = 2 cos x for 0° ≤ x < 360°.

Open this problem in the tool
Show solution
1. Expand the left side:
  
  $\sin x\cos30^{\circ}+\cos x\sin30^{\circ}=\tfrac{\sqrt{3}}{2}\sin x+\tfrac{1}{2}\cos x$
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.
  
  $\tfrac{\sqrt{3}}{2}\sin x=\tfrac{3}{2}\cos x\ \Rightarrow\ \tan x=\sqrt{3}$
3. Tangent repeats every 180°:
  
  $x=60^{\circ},\qquad x=60^{\circ}+180^{\circ}=240^{\circ}$
Answer

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

$x=60^{\circ},\ 240^{\circ}$

Supplementary problems, answers only

Find the exact value of cos 105°.

Show answer

(√2 − √6)/4 ≈ −0.2588

$\cos105^{\circ}=\frac{\sqrt{2}-\sqrt{6}}{4}$
Find the exact value of tan 105°.

Show answer

−2 − √3 ≈ −3.7321

$\tan105^{\circ}=-2-\sqrt{3}$
sin A = 5/13 and cos B = 3/5, with A and B acute. Find cos(A − B).

Show answer

56/65 ≈ 0.8615

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

Show answer

sin 2x

$\sin3x\cos x-\cos3x\sin x=\sin2x$
tan A = 3 and tan B = 1/2, with A and B acute. Find A − B.

Show answer

A − B = 45°

$A-B=45^{\circ}$
Solve cos(x − 60°) = sin x for 0° ≤ x < 360°.

Show answer

x = 75° or x = 255°

$x=75^{\circ},\ 255^{\circ}$

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.

Compound-Angle Formulae

Where the formula comes from

Supplementary problems, answers only

The formulae are a rotation, written out 🖖

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

The tangent formula can fail where the tangent exists 🖖

The formulae are a rotation, written out

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

The tangent formula can fail where the tangent exists