The idea: sin(a + b) from two stacked triangles

A common mistake is to write sin(a + b) = sin a + sin b. Try a = b = 45°: the left side is sin 90° = 1, the right side is 1.414. The real rule, the angle addition identity, is

sin(a + b) = sin a · cos b + cos a · sin b
cos(a + b) = cos a · cos b − sin a · sin b

Both come out of one picture. Draw a segment OP of length 1 at the angle a + b. Its end P has height sin(a + b) and x-coordinate cos(a + b), as on the unit circle. Now split the angle: a right triangle with angle b stands on top of a right triangle with angle a, and a rectangle closes the picture around them.

The rectangle proof with a = 30 degrees and b = 20 degrees. OP = 1 at 50 degrees. Q is the foot of the perpendicular from P onto the ray at angle a, with OQ = cos b and QP = sin b. Below Q the blue triangle ORQ has OR = cos a cos b and RQ = sin a cos b. Above Q the orange triangle QSP has QS = cos a sin b and SP = sin a sin b. The left edge is sin(a + b) and the top piece from the y-axis to P is cos(a + b)
Three right triangles in one rectangle: the right edge is RQ + QS = sin(a + b), and P's x is OR − SP = cos(a + b).

Press Prove to build it one triangle at a time; the Numbers panel measures every side.

Three right triangles

Every side comes from one rule: in a right triangle with hypotenuse h and angle θ, the side next to θ is h · cos θ and the side across is h · sin θ.

TriangleHypotenuseAngleSide next to itSide across
OQP (middle)OP = 1b at OOQ = cos bQP = sin b
ORQ (bottom)OQ = cos ba at OOR = cos a · cos bRQ = sin a · cos b
QSP (top)QP = sin ba at QQS = cos a · sin bSP = sin a · sin b

Why the angle at Q is a

The angle a at O sits between the x-axis and OQ. At Q, the side QS is vertical, so it is at 90° to the x-axis, and QP is at 90° to OQ by construction. Turn both sides of an angle by 90° and the angle between them stays the same, so the angle between QS and QP is a again. This is the one step of the proof that is not just reading off a triangle.

Heights add: the sine formula

P's height is sin(a + b), and that is also the height of the rectangle. The right edge of the rectangle runs from R up to S and is cut at Q into two pieces, one from each small triangle:

sin(a + b) = RQ + QS = sin a · cos b + cos a · sin b

With a = 30°, b = 20°: 0.5 · 0.940 + 0.866 · 0.342 = 0.470 + 0.296 = 0.766 = sin 50°. (Demo: a = 30°, b = 20°)

Widths subtract: the cosine formula

P's x is cos(a + b). The bottom OR reaches all the way to below S, but along the top edge P stops short of S by SP. So

cos(a + b) = OR − SP = cos a · cos b − sin a · sin b

That is why the cosine formula has a minus sign and the sine formula a plus: the two pieces of the right edge stack, but the top piece of the upper triangle points back to the left.

When a + b reaches and passes 90°

At a = b = 45°, P sits right above O: SP = sin² 45° = 0.5 = OR, so cos 90° = 0 and the whole height is 0.5 + 0.5 = 1. (Demo: a + b = 90°) With a = 60°, b = 45° the top piece SP = 0.612 is longer than OR = 0.354, P lands left of O, and cos 105° = −0.259 comes out negative on its own. The picture needs no change: the same three triangles prove the formula for any a and b between 0° and 90°. (Demo: a + b past 90°, cos goes negative)

Double angles: put b = a

With b = a the two pieces of the right edge are both sin a · cos a, so the formulas turn into the double-angle identities:

sin 2a = 2 sin a cos a
cos 2a = cos² a − sin² a = 2 cos² a − 1 = 1 − 2 sin² a

The last two forms use sin² a + cos² a = 1. (Demo: double angle, a = b = 30°: sin 60° = 2 · 0.5 · 0.866 = 0.866, cos 60° = 0.75 − 0.25 = 0.5.)

Left: the rectangle with a = b = 30 degrees; the right edge is cut at Q into two equal blue and orange pieces, each sin a cos a = 0.433, together sin 60 degrees = 0.866. Right: the three forms of cos 2a, cos squared minus sin squared, 2 cos squared minus 1, and 1 minus 2 sin squared, each equal to 0.5 at a = 30 degrees
With a = b the right edge splits into two equal pieces: sin 2a = 2 sin a cos a.

Differences and tan(a + b)

Replace b by −b. Cosine is even and sine is odd (cos(−b) = cos b, sin(−b) = −sin b), so

sin(a − b) = sin a · cos b − cos a · sin b
cos(a − b) = cos a · cos b + sin a · sin b

Divide the sine formula by the cosine formula, then top and bottom by cos a · cos b: tan(a + b) = (tan a + tan b) / (1 − tan a · tan b). The same pair of formulas is what a rotation matrix does when you turn by a and then by b, and what multiplying two complex numbers on the unit circle does: eia · eib = ei(a+b).

See also Proofs Without Words for other identities read off a picture.

What the page leaves out

The picture only works for a, b between 0° and 90°; the formulas hold for every angle, which the rotation or complex-number proof shows in one line. The page also leaves out the half-angle and triple-angle formulas, the product-to-sum formulas (sin a · cos b = ½[sin(a + b) + sin(a − b)]), and how these identities give the derivative of sin x.