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Appendix -Ptolemy's t!Jcorcm tllld t!Je addition f'ormu/us
137
and to recall that
cos (rr +a) =-cos a.
(2)
We use these results to prove Ptolemy's identity.
We want to show that if a + f3 + y + 8 = rr, then
sin a sin{J +sin y sin8 = sin({J + y) sin(a + y).
The right side is a product of two sines. We use (1) to convert this to a sum
of cosines:
sin({J + y) sin(a + y) = ![cos((fJ + y)- (a+ y)}
- cos((fJ + y) +(a+ y))]
=
Hcos ({3 -
a) - cos (a + f3 + 2 y)] .
We must do something about the expression a+ f3 + 2y. We have a+ f3 +
2y =a+ f3 + y + 8 + y- 8 = ;rr + y- 8, and by (2), cos(a+ f3 + 2y) =
cos(rr + y - 8) = - cos(y - 8). So if a + f3 + y + 8 = rr, we have
sin(a + {3) sin(a + 8) = ~ [cos({J -a) + cos(y - 8)].
Since the cosine is an even function, we can write this as
sin(a + {3) sin(a + 8) = ~ [cos(a - {3) + cos(y - 8)] .
Now let us look at the left side of Ptolemy's identity. We have
sina sin{J +sin y sin8 = !(cos(a- {3)- cos(a +
f3)]
+ !(cos(y- 8)- cos(y + 8)].
Nowifx+y = rr, weknowthatcosy = -cosx.Here, (a+f3)+(y+8) =
rr, so cos(y + 8) = - cos(a + {3), and we can write
a
sin sin f3 +sin y sin 8 =
Hcos(a - {3) - cos(a + {3)]
+ Hcos(y - 8) + cos(a + {3)]
= !(cos(a- {3) + cos(y- 8)].
But this is the same expression that we found equal to the left side. So
Ptolemy's identity follows from the formulas for sine and cosine.
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