Is sin(α + β) equal to sin α + sin β?

No, and this is the single most common mistake in trigonometry. A counterexample: sin(30° + 60°) = sin 90° = 1, while sin 30° + sin 60° = 0.5 + 0.866 = 1.366. The correct formula is sin(α + β) = sin α cos β + cos α sin β — four factors on the right, not two terms.

What are the angle sum and difference formulas?

sin(α ± β) = sin α cos β ± cos α sin β and cos(α ± β) = cos α cos β ∓ sin α sin β. In the sine the sign on the right matches the one on the left; in the cosine it is reversed. That is the only thing about them worth memorising.

Where do the double-angle formulas come from?

They are the sum formulas with β = α. Substituting gives sin 2α = 2 sin α cos α and cos 2α = cos²α − sin²α. Using the Pythagorean identity, the second one can also be written as 1 − 2sin²α and as 2cos²α − 1.

Which form of cos 2α should I use?

The one that matches the data. If you know the sine, use 1 − 2sin²α; if the cosine, 2cos²α − 1; if both, cos²α − sin²α is shortest. All three give the same number, so the choice is purely about convenience.

How do I find sin 75° without a calculator?

Split 75° into a sum of table angles: 75° = 45° + 30°. Then sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4 ≈ 0.966. Every angle that is a sum or difference of 30°, 45° and 60° works the same way.

How do I prove a trigonometric identity?

Transform ONE side until it becomes the other — never both at once, and never by moving terms across the equals sign, because the identity is what you are trying to establish. In practice start from the more complicated side, expand the angle sums and doubled angles, and finish by collecting with sin²α + cos²α = 1.

What is tan(α + β)?

tan(α + β) = (tan α + tan β) / (1 − tan α · tan β). It comes from dividing the sine of the sum by the cosine of the sum and cancelling cos α cos β. Watch the denominator: when tan α · tan β = 1 it vanishes and the tangent of the sum does not exist — the two angles then add to 90°.

Do these formulas work in radians?

Yes, unchanged — they are identities between the values of the functions, not between measures of angle. sin(π/4 + π/6) expands exactly as sin(45° + 30°) does and gives the same number, (√6 + √2)/4.

Advanced level

Angle sum and difference

The sine of a sum is not the sum of the sines. See where sin(α ± β) and cos(α ± β) come from, how the double-angle formulas fall out of them, how to get the exact value of sin 75°, and how to use them to prove trigonometric identities.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Reduction formulasReduction formulas bring the sine, cosine and tangent of any angle back to an acute one. Instead of memorising a dozen identities, see the single rule behind them: at 180° and 360° the function stays, at 90° and 270° it swaps for its cofunction, and the sign comes from the quadrant.
The Pythagorean identitysin²α + cos²α = 1 is the Pythagorean theorem written on the unit circle. See where it comes from, how to find the cosine from the sine, and why an answer is incomplete without knowing the quadrant.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Projectile range
The range of a projectile launched at speed v and angle α is R = v² · sin 2α / g — a formula that comes straight out of the double-angle identity. For v = 20 m/s and α = 30° that is 400 · sin 60° / 9.81 = 35.3 m, and for α = 45° it is 40.8 m. Since sin 2α peaks at 2α = 90°, the best launch angle is always 45° — which is why a shot put flies furthest at exactly that angle.
AM radio
Amplitude modulation multiplies a 1000 kHz carrier by a 5 kHz audio signal, and the sum and difference formulas turn that product into a sum: cos A · cos B = ½[cos(A − B) + cos(A + B)]. Instead of one frequency the transmitter emits three — 995, 1000 and 1005 kHz — and that is where a 10 kHz channel width comes from.
Convert it here: Frequency
Tuning an instrument
Two strings sounding 440 Hz and 442 Hz add up to something the sum-to-product formula turns into a single tone whose loudness pulses at 442 − 440 = 2 Hz. The tuner counts those beats — two a second — and tightens the string until they slow to nothing. The same calculation explains the two-tone hum of an unbalanced fan.
A robot arm
A two-link arm of 40 cm and 30 cm, set at 25° and 40°, puts the gripper at x = 40 · cos 25° + 30 · cos(25° + 40°) = 36.3 + 12.7 = 49.0 cm from the base, at a height of y = 40 · sin 25° + 30 · sin 65° = 16.9 + 27.2 = 44.1 cm. The second link is computed from the SUM of two joint angles — without the sum formula there is no way to split that back into the individual joints.
Solar panels
A panel produces power in proportion to the cosine of the angle between the sunlight and the panel normal. A panel tilted 35° from horizontal, with the sun 20° above the horizon, runs at cos(90° − 20° − 35°) = cos 35° = 0.82 of its rated power. Expanding that through cos(α − β) separates what the tilt contributes from what the sun does — which is exactly what a tracker optimises during the day.

All formulas

Sine of a sum
$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
the sign on the right matches the one on the left
Sine of a difference
$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
the same formula with β replaced by −β
Cosine of a sum
$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
careful: the sign on the right is REVERSED
Cosine of a difference
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
the starting formula — the other three follow from it
Tangent of a sum
$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$
the quotient of the two formulas above
Sine of a doubled angle
$\sin 2\alpha = 2 \sin \alpha \cos \alpha$
the sum formula with β = α
Cosine of a doubled angle
$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1$
three forms of one value — pick the one that fits the data

The temptation is strong: if $(a + b)^2$ expands by a binomial rule, surely $\sin(\alpha + \beta)$ expands to $\sin \alpha + \sin \beta$ ? It does not. Just substitute:

\sin(30^\circ + 60^\circ) = \sin 90^\circ = 1, \qquad \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} \approx 1.366

Trigonometric functions are not linear. The real formula looks different — and it opens up everything that follows.

The starting formula: cosine of a difference

Start from one formula; the rest will fall out of it. Take two points on the unit circle: $P$ at angle $\alpha$ and $Q$ at angle $\beta$ .

The chord PQ depends only on the angle between the arms. Turning the whole drawing by −β leaves it unchanged.

Their coordinates are $P = (\cos \alpha, \sin \alpha)$ and $Q = (\cos \beta, \sin \beta)$ . The squared length of the chord comes from the distance between two points:

|PQ|^2 = (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2

Expanding the squares and applying the Pythagorean identity twice:

|PQ|^2 = 2 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)

Now turn the whole drawing through $-\beta$ . The chord is a rigid segment, so its length does not change — while the points move to angles $\alpha - \beta$ and $0$ :

|PQ|^2 = (\cos(\alpha - \beta) - 1)^2 + \sin^2(\alpha - \beta) = 2 - 2\cos(\alpha - \beta)

Comparing the two expressions and cancelling gives the formula:

\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

The other three formulas

They need no separate proof — the reduction formulas from the previous lesson are enough.

Cosine of a sum. Substitute $-\beta$ for $\beta$ and use that the cosine is even and the sine odd:

\cos(\alpha + \beta) = \cos \alpha \cos(-\beta) + \sin \alpha \sin(-\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta

Sine of a sum. Rewrite the sine as the cosine of the complement and regroup the angles:

\sin(\alpha + \beta) = \cos\left(90^\circ - (\alpha + \beta)\right) = \cos\left((90^\circ - \alpha) - \beta\right)

= \cos(90^\circ - \alpha)\cos \beta + \sin(90^\circ - \alpha)\sin \beta = \sin \alpha \cos \beta + \cos \alpha \sin \beta

Sine of a difference follows again by substituting $-\beta$ . The four formulas together:

\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta

\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta

The only thing to memorise is what the sign does: in the sine the right-hand sign matches the left, in the cosine it is reversed.

Tangent of a sum

Divide the sine of the sum by the cosine of the sum, then divide numerator and denominator by $\cos \alpha \cos \beta$ :

\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}

The denominator vanishes exactly when $\tan \alpha \tan \beta = 1$ , that is when $\alpha + \beta = 90^\circ$ — and there the tangent really does not exist. The formula polices its own domain.

Exact values of new angles

The formulas take you past the $30^\circ$ , $45^\circ$ , $60^\circ$ table — just write the angle as a sum or a difference.

Find the exact values of sin 75° and cos 15°.

The same formulas are read backwards as well — and that is the direction exam questions ask in. An expression shaped like the right-hand side collapses to a single function:

\sin 25^\circ \cos 35^\circ + \cos 25^\circ \sin 35^\circ = \sin(25^\circ + 35^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}

The doubled angle

Just put $\beta = \alpha$ into both sum formulas.

The arm of 2α is the arm of α turned through α once more. Its coordinates are expressed through those of the first.

\sin 2\alpha = 2 \sin \alpha \cos \alpha

\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha

The second one has two more forms. Substituting $\cos^2 \alpha = 1 - \sin^2 \alpha$ or $\sin^2 \alpha = 1 - \cos^2 \alpha$ :

\cos 2\alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1

These are not three different formulas but one value written three ways — pick whichever matches the data of the problem.

Given sin α = 3/5 with α acute, find sin 2α and cos 2α.

For the tangent the same substitution gives:

\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha}

Proving identities

A trigonometric identity is an equality true for every angle in its domain. Proving one follows a single rule: transform one side until it becomes the other. You may not move terms across the equals sign or multiply both sides — that would assume the equality already holds, which is exactly what is to be shown.

In practice start from the more complicated side, expand the angle sums and doubled angles, and collect the result with the Pythagorean identity at the end.

Show that (sin α + cos α)² = 1 + sin 2α.

Show that sin 3α = 3 sin α − 4 sin³ α.

Exercises

In the value questions, recognise the formula and collapse the expression to a single function — the answer is an exact value such as √3/2 or 1/2. In the double-angle questions the answer is a fraction, sign included (e.g. −7/25).

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

sin 49° · cos 11° + cos 49° · sin 11° =

Common mistakes

$\sin(\alpha + \beta) = \sin \alpha + \sin \beta$ — the most common error in trigonometry. The counterexample at $30^\circ$ and $60^\circ$ refutes it in two lines.
The wrong sign in the cosine — the cosine of a sum carries a minus on the right, the cosine of a difference a plus. Opposite to the left-hand side, and opposite to the sine.
$\cos 2\alpha = 2\cos \alpha$ — a doubled angle is not a doubled value. At $\alpha = 60^\circ$ , $\cos 120^\circ = -\tfrac{1}{2}$ while $2\cos 60^\circ = 1$ .
Losing the 2 in $\sin 2\alpha$ — the formula is $2\sin \alpha \cos \alpha$ , not $\sin \alpha \cos \alpha$ .
Proving "from both sides at once" — transform one side. Multiplying both sides assumes what is still to be shown.
Splitting an angle into a sum whose parts are not in the table — $75^\circ = 45^\circ + 30^\circ$ works; $75^\circ = 50^\circ + 25^\circ$ buys nothing, because those values are unknown too.

Formula card

Topic: Angle sum and difference

Sine of a sum
$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
the sign on the right matches the one on the left
Sine of a difference
$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
the same formula with β replaced by −β
Cosine of a sum
$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
careful: the sign on the right is REVERSED
Cosine of a difference
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
the starting formula — the other three follow from it
Tangent of a sum
$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$
the quotient of the two formulas above
Sine of a doubled angle
$\sin 2\alpha = 2 \sin \alpha \cos \alpha$
the sum formula with β = α
Cosine of a doubled angle
$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1$
three forms of one value — pick the one that fits the data

The derivation: P and Q sit on the unit circle at angles α and β. The chord PQ depends only on the difference α − β — and that is what gives the formula for cos(α − β).

The doubled angle: the arm of 2α is the arm of α turned through α once more. Its coordinates — cos 2α and sin 2α — are expressed through the coordinates of the first arm.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min