proof of angle sum identities

We will derive the angle sum identities for the various trigonometric functionsDlmfMathworldPlanetmath here. We begin by deriving the identity for the sine by means of a geometric argument and then obtain the remaining identities by algebraic manipulation.

Theorem 1.

Let us make the restrictions 0<x<90 and 0<y<90 for the time being. Then we may draw a triangle ABC such that CAB=x and ABF=y:


Since the angles of a triangle add up to 180, we must have BCA=180-x-y, so we have sin(BCA)=sin(180-x-y)=sin(x+y).

We now draw perpendicularsPlanetmathPlanetmathPlanetmath two different ways in order to derive ratios. First, we drop a perpendicular AD from C to AB:


Since ACD and BCD are right triangles we have, by definition,

cot(CAB)=AD¯/CD¯  cot(ABC)=BD¯/CD¯  sin(CAB)=CD¯/AC¯.

Second, we draw a perpendicular AE form A to BC. Depending on whether x+y<90 or x+y<90 the point E will or will not lie between B and C, as illustrated below. (There is also the case x+y=90, but it is trivial.)

 ???????????????? ΥΥΥΥΥΥΥ ¨¨¨¨¨¨¨¨¨ ABCE
 ?????????????? ¨¨¨¨¨¨¨ wwwwwwwwwwwwwwwwABCE

Either way, ABE and ACE are right triangles, and we have, by definition,

sin(BCA)=AE¯/AC¯  sin(ABC)=AE¯/AB¯.

Combining these ratios, we find that


To finish deriving the sum identity, we manipulate the ratios derived above algebraically and use the fact that AD¯+BD¯=AB¯:

sin(x+y)=sin(BCA) =AB¯sin(ABC)/AC¯

To lift the restriction on the range of x and y, we use the identities for complements and negatives of angles.

Entry under construction

Title proof of angle sum identities
Canonical name ProofOfAngleSumIdentities
Date of creation 2013-05-28 14:35:29
Last modified on 2013-05-28 14:35:29
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 15
Author rspuzio (6075)
Entry type Proof
Classification msc 43-00
Classification msc 51-00
Classification msc 42-00
Classification msc 33B10