One Trigonometric Formula and Its Consequences

The formula below that holds for any angles α, β γ has a two step proof and great many consequences:

(*)

sinα + sinβ + sinγ - sin(α + β + γ)   =

4 sin α + β 2 sin β + γ 2 sin γ + α 2

Proof

Copyright © 1996-2009 Alexander Bogomolny

From the Addition formulas for sine and cosine one can easily get important formulas for the sum and difference of two sines or of two cosines:

(1)	sinα + sinβ   =	2 sin α + β 2 cos α - β 2
(2)	sinα - sinβ   =	2 sin α - β 2 cos α + β 2
(3)	cosα + cosβ   =	2 cos α + β 2 cos β - α 2
(4)	cosα - cosβ   =	2 sin α + β 2 sin β - α 2

For example, since

then

Now (1) follows by setting α = a + b and β = a - b and solving the system for a and b:

To prove (*),

(*)

sinα + sinβ + sinγ - sin(α + β + γ)   =

4 sin α + β 2 sin β + γ 2 sin γ + α 2

we apply (1) and (2) to the left side and subsequently (4) to the result:

as required.

When α, β, γ are angles of a triangle, i.e., when, in particular, α + β + γ = π, we may observe that, for example,

which allows to reduces (*) to

because sin (π/2 - x) = cos(x).

The latter identity (**) has further consequences. Right now, I'll give a second, more direct proof of (**) [Stanford Problem Book, #47.4]. It's not easier than the above and I do not know if it admits a simpler proof than the more general (*).

Let α = 2u, β = 2v, γ = π - 2u - 2v. The, using the formula for the double argument [sin(2x) = 2 sin(x)cos(x)] (**) is transformed into

which is actually (1) and hence true.

Reference

1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009

Copyright © 1996-2009 Alexander Bogomolny