Multiple Angles

Take any angle A and double it. What is \sin 2A in terms of \sin A and \cos A?

A concrete case first. If A = 30°, then \sin 2A = \sin 60° = \sqrt{3}/2. And \sin 30° = 1/2. Doubling the angle does not double the sine — \sin 60° is not 2 \times \sin 30°. So the relationship between \sin A and \sin 2A is not simple multiplication. Something more interesting is happening.

You could reach for a calculator, but that only gives you a decimal for one specific angle. The real question is structural: is there a formula that works for every angle A, expressed purely in terms of the trigonometric ratios of A itself?

There is, and if you already know the compound angle formulas, it takes one line to find it. That is the pattern of this entire article: every formula here is a special case of compound angles. You are not memorising new identities — you are seeing familiar ones from a new perspective.

Double angle formulas

sin 2A

Start with the compound angle formula for sine:

Set B = A:

Why: the two terms are identical — both are \sin A \cos A — so they add up to twice that.

That is the double angle formula for sine. One line, straight from compound angles.

cos 2A

The same move, applied to cosine:

Set B = A:

Why: the compound angle formula gives a product minus a product, and when both angles are the same, each product becomes a square.

This formula has three equivalent forms, because \sin^2 A + \cos^2 A = 1 lets you eliminate either \sin^2 A or \cos^2 A:

Form 1: \cos 2A = \cos^2 A - \sin^2 A (the original)

Form 2: Replace \sin^2 A with 1 - \cos^2 A:

Form 3: Replace \cos^2 A with 1 - \sin^2 A:

All three are the same identity wearing different clothes. Which one you use depends on what the problem gives you. If you only know \cos A, use Form 2. If you only know \sin A, use Form 3. If you know both, Form 1 is cleanest.

tan 2A

From the compound angle formula for tangent:

Set B = A:

This formula blows up when \tan^2 A = 1, i.e. when A = 45° or A = 135° — which is correct, because \tan 90° and \tan 270° are undefined.

All double angle formulas, collected

Expressing sin 2A and cos 2A in terms of tan A alone

There is a useful variant: express double angle formulas entirely in terms of \tan A. Start with \sin 2A = 2\sin A \cos A. Divide numerator and denominator by \cos^2 A:

Why: dividing top and bottom by \cos^2 A turns every \sin A into \tan A and every \cos A into 1, while the denominator becomes 1 + \tan^2 A = \sec^2 A.

Similarly:

These forms are useful when \tan A is the only ratio you have, and they connect to the t-substitution covered in the going-deeper section below.

A visual check

The formula \sin 2A = 2\sin A \cos A says that doubling the angle does not double the sine. Instead, the relationship involves a product of sine and cosine.

Triple angle formulas

Set B = 2A in the compound angle formula to express \sin 3A and \cos 3A in terms of A alone.

sin 3A

Substitute \sin 2A = 2\sin A \cos A and \cos 2A = 1 - 2\sin^2 A:

Why this form of \cos 2A? Because the goal is to express everything in terms of \sin A. Using \cos 2A = 1 - 2\sin^2 A gives that; any other form would leave \cos A terms that need further conversion.

Replace \cos^2 A with 1 - \sin^2 A:

cos 3A

Substitute \cos 2A = 2\cos^2 A - 1 and \sin 2A = 2\sin A \cos A:

Replace \sin^2 A with 1 - \cos^2 A:

tan 3A

Apply the tangent addition formula with A and 2A:

Substitute \tan 2A = \frac{2\tan A}{1 - \tan^2 A} and let t = \tan A for brevity:

Sanity check. Verify \sin 3A = 3\sin A - 4\sin^3 A at A = 30°:

Left side: \sin 90° = 1.

Right side: 3 \cdot \frac{1}{2} - 4 \cdot \frac{1}{8} = \frac{3}{2} - \frac{1}{2} = 1. Correct.

And \cos 3A = 4\cos^3 A - 3\cos A at A = 30°:

Left side: \cos 90° = 0.

Right side: 4 \cdot \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \cdot \frac{\sqrt{3}}{2} = 4 \cdot \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0. Correct.

The coefficient pattern — 3 and 4 appearing in both sine and cosine, but with different signs — is not a coincidence. It comes from the binomial expansion of (\cos A + i\sin A)^3 via De Moivre's theorem. For higher multiples (4A, 5A, ...), the pattern continues with binomial coefficients.

A useful observation: \sin 3A = 0 when 3\sin A - 4\sin^3 A = 0, i.e. when \sin A(3 - 4\sin^2 A) = 0. This factors as \sin A = 0 or \sin^2 A = 3/4, i.e. \sin A = \pm\sqrt{3}/2. So 3A is a multiple of 180° exactly when A = 0°, 60°, 120°, 180°, \ldots — which is geometrically obvious but algebraically satisfying to confirm.

Half angle formulas

The double angle formulas run in one direction: from A to 2A. Reverse them, and you go from A down to A/2. The half angle formulas are the double angle formulas, read backwards.

Why would you want formulas for half an angle? One reason: you know the trigonometric ratios of 45°, but what about 22.5°? That is 45°/2. You know \cos 60° = 1/2, but what about \cos 30°? That is 60°/2. Half angle formulas let you compute the exact values of new angles from angles you already know — extending your table of known values in a different direction from compound angles.

Deriving sin(A/2)

Start with \cos 2\theta = 1 - 2\sin^2\theta (Form 3 of the double angle formula for cosine). Solve for \sin^2\theta:

Now replace \theta with A/2, so that 2\theta = A:

Why the \pm? Because the square root could be positive or negative — the sign depends on which quadrant A/2 falls in. If A/2 is in the first or second quadrant, take the positive root. If it is in the third or fourth, take the negative.

Deriving cos(A/2)

Start with \cos 2\theta = 2\cos^2\theta - 1 (Form 2). Solve for \cos^2\theta:

Replace \theta with A/2:

Deriving tan(A/2)

Divide:

There is a cleaner form that avoids the \pm ambiguity. Multiply numerator and denominator inside the square root by (1 - \cos A):

Why does this work? The numerator comes from \sqrt{(1 - \cos A)^2} = |1 - \cos A| = 1 - \cos A (since 1 - \cos A \geq 0 always). The denominator comes from \sqrt{(1 + \cos A)(1 - \cos A)} = \sqrt{1 - \cos^2 A} = |\sin A|. The signs of \sin A and \tan(A/2) agree when A is between 0 and 2\pi, so the formula works without a \pm.

Alternatively, multiply by (1 + \cos A) instead:

Both forms are useful. The first is better when \cos A is known but \sin A is not. The second is better in the opposite case.

Power reduction formulas

The half angle derivation passed through an intermediate step that is important in its own right. Rearranging the double angle formulas gives you a way to express \sin^2 A and \cos^2 A in terms of \cos 2A — reducing the power from 2 down to 1.

These are the same equations used to derive the half angle formulas, just without the substitution \theta = A/2. They are indispensable in integration (when you study calculus, every integral of \sin^2 x or \cos^2 x uses exactly these formulas) and in simplifying expressions that involve even powers of trigonometric functions.

Worked examples

Common confusions

• "\sin 2A = 2\sin A." False. This is the same linearity error from compound angles, reappearing. The correct formula is \sin 2A = 2\sin A \cos A — there is a \cos A factor that people drop. At A = 30°: \sin 60° = \sqrt{3}/2 \approx 0.866, but 2\sin 30° = 1. They are not equal.

• "\cos 2A has only one form." It has three: \cos^2 A - \sin^2 A, 2\cos^2 A - 1, and 1 - 2\sin^2 A. All three are the same identity. Picking the wrong one doesn't give a wrong answer, but it gives a harder computation. Choose the form that matches what you know about A.

• Forgetting the \pm in half angle formulas. The formulas \sin(A/2) = \pm\sqrt{\cdots} have a sign ambiguity that must be resolved by looking at which quadrant A/2 is in. Many students write the positive root automatically, which is correct only when A/2 is in the first or second quadrant.

• "\tan 2A = 2\tan A." Also false. The denominator 1 - \tan^2 A is doing real work. For example, at A = 30°: \tan 60° = \sqrt{3} \approx 1.732, but 2\tan 30° = 2/\sqrt{3} \approx 1.155. The difference is significant.

• Confusing power reduction with half angle. The formulas \sin^2 A = (1 - \cos 2A)/2 and \sin(A/2) = \pm\sqrt{(1 - \cos A)/2} look similar but do different things. Power reduction takes \sin^2 A and turns it into \cos 2A. The half angle formula takes \sin(A/2) and turns it into \cos A. They come from the same equation, but the substitution is different.

Going deeper

The t-substitution

Let t = \tan(A/2). Then every trigonometric function of A can be written as a rational function of t:

Here is why. Start with \sin A = 2\sin(A/2)\cos(A/2). Divide numerator and denominator by \cos^2(A/2):

The cosine derivation is similar, starting from \cos A = \cos^2(A/2) - \sin^2(A/2) and dividing by \cos^2(A/2) + \sin^2(A/2).

This substitution — called the Weierstrass substitution or t-substitution — is a powerful tool in calculus. It converts any integral involving trigonometric functions into an integral involving a rational function of t, which can always be computed (in principle) using partial fractions.

Chebyshev polynomials

The triple angle formulas \sin 3A = 3\sin A - 4\sin^3 A and \cos 3A = 4\cos^3 A - 3\cos A express \sin 3A and \cos 3A as cubic polynomials in \sin A and \cos A respectively. This pattern continues: \cos nA can always be written as a polynomial of degree n in \cos A. These polynomials are called Chebyshev polynomials (of the first kind), and they appear in numerical analysis, approximation theory, and physics.

The first few are:

Each row can be computed from the previous two using the recurrence T_{n+1}(c) = 2c \cdot T_n(c) - T_{n-1}(c), which itself follows from the compound angle identity \cos(n+1)A = 2\cos A \cos nA - \cos(n-1)A.

Visualising cos 2A: three forms, one curve

It is worth seeing that the three forms of \cos 2A really do agree everywhere.

The Kerala school and infinite series

The multiple angle formulas played a role in one of the most remarkable achievements of Indian mathematics. In the 14th century, Madhava of the Kerala school used relationships between \sin A and \sin nA (for large n) to derive infinite series for \sin A and \cos A:

These are the same Taylor series that would be independently discovered in Europe two centuries later. The path from multiple angle formulas to infinite series is not obvious — it involves letting the number of angle subdivisions grow without bound — but it shows how deeply the formulas in this article are connected to the foundations of calculus.

Where this leads next

Multiple angle formulas sit at a crossroads. In one direction, they combine with each other to produce the transformation formulas. In another, they lead to equations and identities. In a third, they generalise to complex exponents.

• Transformation Formulas — add and subtract double angle formulas to convert sums into products and products into sums.
• Trigonometric Equations — many equations, like \sin 2x = \sin x, are solved by expanding the multiple angle and factoring.
• Compound Angles — the foundation that everything in this article rests on. Go back here if any derivation above felt unclear.
• Trigonometric Identities — the broader toolkit, including Pythagorean identities and reciprocal relations, that these formulas fit into.
• De Moivre's Theorem — the complex-number framework that generates all multiple angle formulas at once, for any integer n.