The Unit Circle

Chapter 4

Trigonometric identities

Every identity you need is a fact about one picture. Here are the pictures, with the squares and rectangles you can drag.

8 min read · 4 interactive figures

Drag the point. The blue square (cos²θ) and the vermilion square (sin²θ) always add up to the square on the radius, 1.

A trigonometric identity is an equation that holds at every angle, such as sin²θ + cos²θ = 1. The few you need most come from two pictures. The Pythagorean identities are Pythagoras’s theorem on the unit circle. The angle-sum identities, and the double-angle identities that follow from them, come from boxing two right triangles in a rectangle. Once you have seen the pictures, the formulas stop being a list and become consequences[1][2].

The Pythagorean identities

The point (cos θ, sin θ) is on the unit circle, so its coordinates satisfy x² + y² = 1. That is the first identity. Oliver Byrne’s 1847 edition of Euclid replaced the letters of geometry with diagrams in which each line, angle and polygon is printed in a primary color[3]; drawn his way, Pythagoras’s theorem becomes a statement about colored squares, and the colors on this site follow him.

Figure 1 Drag the point. The blue square has area cos²θ, the vermilion square sin²θ; together they always match the square on the radius, area 1. The bars underneath add the areas. Switch the triangle to see the second identity.#
sin²θ + cos²θ = 1     1 + tan²θ = sec²θ     1 + cot²θ = csc²θ(4.1)

The second and third forms are the first one divided through by cos²θ and by sin²θ. They also have pictures of their own: switch the figure to the triangle with legs 1 and tan θ, which sits on the tangent line from Chapter 1. Its hypotenuse runs from the center to the tangent line, and its length is sec θ. The squares on its sides give 1 + tan²θ = sec²θ directly. These identities fail only where a term is undefined: the tangent form at 90° and 270°, where cos θ = 0.

The angle-sum formulas

Here is where intuition usually goes wrong.

Question 1 A prediction about adding angles.#

To find sin(α + β), draw a radius of length 1 at angle α + β, ending at the point P. From P, drop a perpendicular onto the ray at angle α; call its foot Q. Now there is a right triangle OQP with hypotenuse 1 and angle β at O, so OQ = cos β and QP = sin β. Box the figure in a rectangle with its bottom edge on the x-axis.

Figure 2 Drag Q to change α, or P to change β. The right side of the rectangle is made of a vermilion piece and a yellow piece; together they are the height of P, which is sin(α + β). The top side is made of a blue piece, cos(α + β), and a purple piece.#

Read the rectangle’s right-hand side in two ways. It is the height of P, which is sin(α + β). It is also the sum of two pieces. The lower piece is the height of Q: Q is at distance cos β along a ray at angle α, so its height is sin α cos β. The upper piece is the vertical part of QP. The angle between QP and the vertical is α again (marked at P in the figure: QP is perpendicular to OQ, so it is turned from the vertical by the same angle that OQ is turned from the horizontal), so the vertical part of QP is sin β cos α. Adding the pieces:

sin(α + β) = sin α cos β + cos α sin β(4.2)

The top side gives the cosine formula the same way. Its full width is the x-coordinate of Q, cos α cos β. It is split at P into cos(α + β), the x-coordinate of P, and the horizontal part of QP, which is sin α sin β. So

cos(α + β) = cos α cos β − sin α sin β(4.3)

Replace β by −β, and use cos(−β) = cos β and sin(−β) = −sin β, to get the 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.

These formulas are what fill in the exact values between the special angles of Chapter 1. For example,

sin 15° = sin(45° − 30°) = √22·√32 − √22·12 = √6 − √24 ≈ 0.2588(4.4)

The formulas are old. Ptolemy computed his chord table around AD 150 with relations for chords that are equivalent to our formulas for sin(a + b) and sin(a − b), which he deduced from the theorem about quadrilaterals inscribed in a circle that still carries his name[4]. documented

Double-angle and half-angle

Tick “Make β = α” in the figure and the rectangle proves the double-angle identities, because α + α = 2α:

sin 2α = 2 sin α cos α     cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α(4.5)

The last two forms of cos 2α use the Pythagorean identity to trade sin² for 1 − cos² or the other way around. The sine form was known to Abu’l-Wafa by 980[5]. documented Solving the cosine forms for the squares gives the half-angle identities, which turn a cosine into the sine or cosine of half the angle:

sin²α2 = 1 − cos α2     cos²α2 = 1 + cos α2(4.6)

They give a second route to cos 15°: √((1 + cos 30°)/2) = √(2 + √3)/2. That does not look like (√6 + √2)/4, the value from the difference formula, but square both and you get (2 + √3)/4 each time. Two different-looking exact forms of one number is a common source of confusion in homework answers; squaring, or a calculator, settles it.

How to test an identity

Faced with a candidate identity, there are two jobs: find out whether it is true, then prove it. A computer is very good at the first job. Evaluate both sides at a few thousand angles and compare. If the two sides ever differ by more than rounding error, the identity is false, and the angle where they differ is a counterexample.

Figure 3 Choose a candidate from the list, which mixes true identities with three classic false ones. Top: both sides plotted against the angle from −4π to 4π, the same range the engine tests. Bottom: the gap between them on a logarithmic scale. True identities sit at the floor of 64-bit rounding error at every angle; false ones leave it almost everywhere.#

The three false candidates in the list are the ones students write most. “sin(α + β) = sin α + sin β” treats sine as multiplication by a number. “sin 2θ = 2 sin θ” does the same with a factor. “sin²θ = sin(θ²)” misreads the notation: sin²θ means (sin θ)², the square of the height, and not the sine of a squared angle. Each fails at a single, easy-to-check angle, given in the sentence under the figure when you select it.

Agreement at thousands of angles is strong evidence, but it is not a proof, and the figure says so. An identity is a claim about every angle, infinitely many, and a finite test cannot rule out a disagreement somewhere it did not look. The pictures above are the proofs. The tester is how a mathematician checks a conjecture before trying to prove it, and how this site’s engine checks itself: every true identity on this page is evaluated at 4,096 angles from −4π to 4π in the engine’s test suite, and every false one must fail.

The identities on one page

IdentityWhere it comes from
sin²θ + cos²θ = 1The point is on the unit circle (Figure 1)
1 + tan²θ = sec²θ; 1 + cot²θ = csc²θDivide the first by cos²θ or sin²θ; the tangent-line triangle
tan θ = sin θ/cos θ; sec θ = 1/cos θ; csc θ = 1/sin θSimilar triangles on the circle (Chapter 1)
sin(−θ) = −sin θ; cos(−θ) = cos θReflection in the x-axis
sin(π/2 − θ) = cos θ; cos(π/2 − θ) = sin θReflection in the line y = x swaps the coordinates
sin(α ± β) = sin α cos β ± cos α sin βThe rectangle (Figure 2)
cos(α ± β) = cos α cos β ∓ sin α sin βThe rectangle (Figure 2)
sin 2α = 2 sin α cos αAngle sum with β = α
cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²αAngle sum with β = α, then the Pythagorean identity
sin²(α/2) = (1 − cos α)/2; cos²(α/2) = (1 + cos α)/2Solve the double-angle forms for the squares

Every row is in DLMF §4.21, the standard reference for these formulas[1]. The trig identities cheat sheet adds the product-to-sum, sum-to-product, half-angle and triple-angle formulas on one printable page. The angle-sum formulas also hide the laws of triangles: Chapter 5 uses the circle to solve any triangle, not only right-angled ones.