Compound Angles

You know that \sin 30° = \tfrac{1}{2} and \sin 45° = \tfrac{1}{\sqrt{2}}. What is \sin 75°?

Here is a tempting but wrong answer: \sin 75° = \sin 30° + \sin 45° = \tfrac{1}{2} + \tfrac{1}{\sqrt{2}} \approx 1.207. That cannot be right — no sine value exceeds 1. The sine function is not additive. You cannot split the angle and add the sines.

But 75° = 30° + 45°, and there is a formula for \sin(30° + 45°) — it just looks different from naive addition. That formula is

Plug in A = 30°, B = 45°:

That is a legitimate sine value — it sits between 0 and 1. And it matches the actual value of \sin 75° exactly.

The formula that made this work is called a compound angle formula. "Compound" because the angle is built from two simpler pieces. There are six of them in total — for sin, cos, and tan of both A + B and A - B — and once you have the first one, the rest follow by short, clean algebra. The real question is: where does the first one come from?

Where the formula comes from

The cleanest proof uses nothing more than a right triangle sitting inside another right triangle. Start with two acute angles A and B. Stack them: place angle A at the base, and then lay angle B on top of it. The combined angle is A + B.

Here is the construction in detail. Take a line segment OP of length 1 (the unit hypotenuse), making an angle A + B with the horizontal. Drop a perpendicular from P to the horizontal axis — its length is \sin(A + B). That is the quantity you want to express in terms of A and B separately.

Now draw the line OQ at angle A from the horizontal. Drop a perpendicular from P onto the line OQ, meeting it at R. This creates two right triangles: \triangle OPR (with angle B at O) and a smaller triangle sitting on top of it.

In triangle \triangle OPR:

• The hypotenuse is OP = 1.
• The side PR (perpendicular to OQ) has length \sin B.
• The side OR (along OQ) has length \cos B.

Now project the vertical height. The total vertical drop from P to the x-axis — which is \sin(A+B) — has two parts:

1. The vertical component of OR. Since OR = \cos B and OR makes angle A with the horizontal, its vertical component is \cos B \cdot \sin A.
2. The vertical component of PR. Since PR = \sin B and PR is perpendicular to OQ (which itself makes angle A with horizontal), PR makes angle A with the vertical. So its vertical component is \sin B \cdot \cos A.

Add them:

That is the formula. It did not come from algebra — it came from stacking two right triangles and decomposing a single vertical line into two projections.

The cos(A + B) formula

The same picture gives you cosine. The horizontal distance from O to the foot of the perpendicular from P is \cos(A + B). Decompose it the same way:

1. The horizontal component of OR. Since OR = \cos B and OR makes angle A with the horizontal, its horizontal component is \cos B \cdot \cos A.
2. The horizontal component of PR. Since PR = \sin B and PR makes angle A with the vertical, its horizontal component is \sin B \cdot \sin A. But this component points in the opposite direction — to the left — so it is subtracted.

Put them together:

The minus sign is not arbitrary. It appears because the horizontal projection of PR points in the opposite direction to the horizontal projection of OR. The geometry forces the sign.

The cos(A + B) formula: graphical verification

You can verify the cosine formula with a graph. Plot \cos(A + 30°) for varying A, and also plot \cos A \cos 30° - \sin A \sin 30°. They should be identical.

The subtraction formulas

Once you have the addition formulas, the subtraction versions cost nothing. Replace B with -B everywhere and use two facts: \sin(-B) = -\sin B and \cos(-B) = \cos B.

For sine:

For cosine:

Notice that the sign pattern swaps. In the sine formula, + becomes -. In the cosine formula, - becomes +. This makes a tidy mnemonic: sine keeps the sign, cosine flips it.

All four formulas, collected

The tangent formula

Tangent is sine divided by cosine. So the compound angle formula for tangent should come from dividing the sine formula by the cosine formula. It does.

This expression is correct but unwieldy. To get a cleaner form, divide both the numerator and the denominator by \cos A \cos B:

Why divide by \cos A \cos B? Because every term then simplifies to either 1 or a tangent ratio — and tangent is the ratio you want the answer in.

Similarly, for subtraction:

The denominator 1 - \tan A \tan B is significant. When \tan A \tan B = 1, the denominator is zero, and \tan(A + B) is undefined — meaning A + B is an odd multiple of 90°. For instance, \tan 50° \cdot \tan 40° = 1 because 50° + 40° = 90°.

A quick sanity check

Before using the formulas in anger, verify them against values you already know. This is worth doing — a formula that fails a sanity check is a formula you have remembered wrong.

Test 1. Set B = 0:

Good — adding zero to the angle should change nothing, and the formula confirms that.

Test 2. Set A = B:

That is the double angle formula \sin 2A = 2 \sin A \cos A — which you may have already seen. It falls straight out of the compound angle formula with B = A. That is reassuring: the double angle formulas are special cases of compound angles, not separate formulas.

Test 3. Set A = 45°, B = 45°:

And indeed \cos 90° = 0. The formula works.

Test 4. Set B = 90° - A in the cosine addition formula:

Correct. This also confirms the co-function identities \cos(90° - A) = \sin A and \sin(90° - A) = \cos A.

Worked examples

Common confusions

• "\sin(A + B) = \sin A + \sin B." This is the single most common error in trigonometry. It is false for almost every pair of angles. The sine function is not linear — you cannot distribute it over addition. The correct formula has four terms (two products), not two sums.

• "The formula only works for acute angles." The geometric proof above is drawn with acute angles, but the formulas hold for all real angles. You can verify this using the unit-circle definitions: for any point on the unit circle at angle \theta, \cos \theta is the x-coordinate and \sin \theta is the y-coordinate, and the algebra goes through identically.

• "I can never remember which sign goes where." Here is a pattern. In \sin(A \pm B), the sign between the two terms matches the \pm in the argument. In \cos(A \pm B), the sign between the two terms is opposite to the \pm in the argument. So: sine is "friendly" (same sign), cosine is "contrary" (flips the sign).

• "\tan(A + B) = \tan A + \tan B." Also false, and for the same reason. The correct formula has a denominator 1 - \tan A \tan B that people forget. This denominator is doing real geometric work — it accounts for the fact that tangent can blow up to infinity, which simple addition cannot capture.

• Mixing up the cross-terms in sine. The formula is \sin A \cos B + \cos A \sin B — each term is a sine-cosine pair where the angle switches. Some students write \sin A \sin B + \cos A \cos B, which is actually \cos(A - B), a completely different formula.

Going deeper

A proof using the unit circle and distance

There is a proof of the cosine subtraction formula that avoids triangles entirely and uses only the distance formula on the unit circle.

Place two points on the unit circle: P at angle A with coordinates (\cos A, \sin A), and Q at angle B with coordinates (\cos B, \sin B). The straight-line distance between P and Q can be computed two ways.

By the distance formula:

Expand:

By the cosine rule (applied to the triangle OPQ, where O is the origin and both OP and OQ have length 1):

Set the two expressions equal:

This proves the cosine subtraction formula for all angles at once — not just acute — because the distance formula and the cosine rule work for any angles on the unit circle. Every other compound angle formula follows from this one by substitution.

Connection to rotation matrices

In linear algebra, rotating a point (x, y) by angle \theta counterclockwise is done by multiplying by the rotation matrix

When you rotate by A and then by B, the combined rotation is R(A + B) = R(A) \cdot R(B). Multiply the two matrices out:

The entries of this matrix are exactly \cos(A+B) and \sin(A+B). So the compound angle formulas are secretly the statement that two successive rotations are the same as one combined rotation. The algebra of trigonometry is the algebra of rotations. This connection runs deep — it is the beginning of the path toward Euler's formula e^{i\theta} = \cos\theta + i\sin\theta, which you will meet in De Moivre's Theorem.

An application: the tangent of 15° and 75°

From the tangent formula:

Rationalise: multiply top and bottom by (\sqrt{3} - 1):

And \tan 75° = \tan(45° + 30°) = 2 + \sqrt{3}. Notice that \tan 15° \cdot \tan 75° = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1. This is not a coincidence — it holds because 15° + 75° = 90°, and for complementary angles \tan\alpha \cdot \tan(90° - \alpha) = \tan\alpha \cdot \cot\alpha = 1.

Writing a sin θ + b cos θ as a single sinusoidal

One of the most powerful applications of compound angles is rewriting a sum like a\sin\theta + b\cos\theta as a single trigonometric function. This comes up constantly in solving trigonometric equations and in finding maximum/minimum values.

Start with a\sin\theta + b\cos\theta. Factor out \sqrt{a^2 + b^2}:

Now notice that \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2 = 1, so there exists some angle \alpha with \cos\alpha = \frac{a}{\sqrt{a^2 + b^2}} and \sin\alpha = \frac{b}{\sqrt{a^2 + b^2}}.

Substituting:

where \alpha = \arctan(b/a).

This has an immediate consequence: the maximum value of a\sin\theta + b\cos\theta is \sqrt{a^2 + b^2} and its minimum is -\sqrt{a^2 + b^2}. For example, 3\sin\theta + 4\cos\theta ranges between -5 and 5, because \sqrt{9 + 16} = 5. No calculus needed — just compound angles.

Aryabhata and Indian trigonometry

The compound angle concept has deep roots in Indian mathematics. Bhaskara II's 12th-century work Siddhanta Shiromani contains results equivalent to these formulas, used for computing planetary positions. The tradition of building exact values of trigonometric ratios from simpler ones — exactly what the compound angle formulas do — was well established in the Kerala school of mathematics by the 14th century, where Madhava and his successors developed infinite series for sine and cosine that implicitly depended on these identities.

Indian astronomers needed these formulas for a very practical reason: computing the positions of planets at future dates required adding angular displacements, and the trigonometric ratios of the combined displacement had to be expressed in terms of known tabulated values. The compound angle formulas were, in this sense, a computational tool before they were a theorem — and the Indian tradition preserved them as working algorithms alongside the geometric insights.

Where this leads next

The compound angle formulas are the foundation for the entire next layer of trigonometric identities. Every formula in the following articles is a direct consequence of what you have just learned.

• Multiple Angles — set B = A to get double angles, set B = 2A to get triple angles, and halve the angle to get half-angle formulas.
• Transformation Formulas — add and subtract the compound angle formulas to convert sums of sines into products, and products into sums.
• Trigonometric Identities — compound angles are the workhorse behind most identity proofs.
• Trigonometric Equations — many equations can only be solved by first rewriting a sum as a single trigonometric function using these formulas.
• De Moivre's Theorem — the complex-number version of compound angles, where all of trigonometry collapses into the algebra of exponents.