Download HYPERBOLIC TRIGONOMETRY Consider a hyperbolic triangle

HYPERBOLIC TRIGONOMETRY
STANISLAV JABUKA
Consider a hyperbolic triangle ∆ABC with angle measures (∠A)◦ = α, (∠B)◦ = β and
(∠C)◦ = γ. The purpose of this note is to develop formulae that allow for a computation of
the hyperbolic lengths a = d(AB), b = d(AC) and c = d(BC) of the sides of this triangle in
terms of α, β and γ. We do this by pursuing two different approaches. The first one, given in
section 1, uses standard computational techniques from analytic geometry of the Euclidean
plane. The second, found in section 2, uses the correspondence between the Poincaré disk
model and the Beltrami-Klein model of hyperbolic geometry.
1. First approach - Euclidean analytic geometry
Let Γ be the circle centered at the origin O = (0, 0) of a coordiante system in the Euclidean
plane and let r be the radius of Γ. We take Γ as the circle for the Poincaré disk model for
hyperbolic geometry.
Let A, B and C inside of Γ be three points that are non-collinear in the hyperbolic sense
(i.e. A, B and C do not lie on a diameter of Γ nor on a circle perpendicular to Γ). Without
loss of generality we can assume that A = (0, 0) and C = (b, 0) for some choice of 0 < b < r.
This is so since every hyperbolic triangle can be made to have one of its vertices agree with
the center of Γ by an inversion (along a circle perpendicular to Γ). Since such an inversion
preserves the angles and hyperbolic sidelengths of the triangle, we may well replace the
original with the one gotten by its inversion. Moreover, by an additional rotation about O,
we can assume that C lies on the x-axis of the coordinate system. Finally, by a reflection
across the x-axis, we can assume that B lies in the first quadrant of the chosen coordinate
system, see figure 1.
In this section we shall derive a relation between α, β, γ and b by pursuing the following
three step strategy:
(1) Find the center X = (x0 , y0 ) and radius ρ of the circle δ containing B and C and
perpendicular to Γ. Express x0 , y0 , ρ as functions of α, γ, b.
(2) Find the coordinates of the point B = (xB , yB ) by intersecting δ with the line containing AB.
(3) Find the slope of the tangent line to δ at B from knowing the equation of δ and
the coordinates (xB , yB ) of B. Equate this slope with tan(α + β), the latter is an
equation relating α, β, γ and b.
Note that knowing b allows one to compute the hyperbolic length b = d(AC) easily using
the formula
r+b
b = d(AC) = log
r−b
Having outlined our strategy, we start with step 1. All geometric objects used in the subsequent paragraphs can be found in figure 1.
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2
STANISLAV JABUKA
B0
δ
m3
Γ
B
X
m1
A
C0
C
m4
m5
m2
Figure 1. The triangle ∆ABC and the various points and lines associated
with it.
1.1. Step 1. Recall that δ is the unique circle perpendicular to Γ passing through the points
B and C. Let C 0 be the inverse of C in Γ. An easy computation then shows that
2 r
0
,0
C =
b
Since δ ⊥ Γ, the point C 0 must also lie on δ. But then the center X = (x0 , y0 ) of δ must lie
on the perpendicular bisector m1 of CC 0 , i.e. we must have
r 2 + b2
2b
To find y0 we note that the tangent line m2 to δ at C is given by the equation
x0 =
m2
:
y = − tan γ · (x − b)
and so the line m3 passing through C and perpendicular to m2 , is given by
1
m3 : y =
(x − b)
tan γ
Since X must lie on m3 , and having computed its x-coordinate x0 above, we are then in a
position to determined y0 as:
r 2 − b2
y0 =
2b tan γ
HYPERBOLIC TRIGONOMETRY
3
With the coordinates (x0 , y0 ) of X in place, it is now easy to determine the radius ρ of δ as
s
2 2
2
r 2 + b2
r − b2
r 2 − b2
ρ = XB =
−b +
−0 =
2b
2b tan γ
2b sin γ
To summarize, we found that
2
r + b2 r 2 − b2
X=
,
2b
2b tan γ
and
ρ=
r 2 − b2
2b sin γ
and thus the equation of δ reads
2 2 2
2
r 2 + b2
r 2 − b2
r − b2
(1)
x−
+ y−
=
2b
2b tan γ
2b sin γ
From here on we split our computations into two cases according to whether α is 90o or not.
1.2. Step 2 with α 6= 90o . For simplicity we assume in this section that α < 90◦ , the case
of α > 90◦ requires only a few sign changes, a case that is left to the interested reader.
The Euclidean line m4 containing the edge AB has the equation
m4
:
y = tan α · x
The point B is then one of the points gotten by intersecting the circle δ with m4 . For ease
of computation we leave the equation of δ in the form (x − x0 )2 + (y − y0 )2 = ρ2 . We plug
y = tan α · x into the latter and solve for x:
x2 − 2xx0 + x20 + x2 tan2 α − 2xy0 tan α + y02 − ρ2 = 0
x2 (1 + tan2 α) − 2x(x0 + y0 tan α) + (x20 + y02 − ρ2 ) = 0
1
x2 2 − 2x(x0 + y0 tan α) + (x20 + y02 − ρ2 ) = 0
cos α
Solving this quadratic equation for x leads to:
q
cos2 α
2
2
2
2
2
2(x0 + y0 tan α) ± 4(x0 + y0 tan α) − 4(1 + tan α)(x0 + y0 − ρ )
x1,2 =
2
q
2
2
2
2
2
= cos α x0 + y0 tan α ± (ρ/ cos α) − y0 + 2x0 y0 tan α − x0 tan α
p
2
2
2
= cos α x0 + y0 tan α ± (ρ/ cos α) − (y0 − x0 tan α)
Since both the point B and its inverse B 0 must lie on δ (since δ ⊥ Γ) but one of them,
namely B, lies inside of Γ, the other, i.e. B 0 , must lie outside of Γ. Since α is assumed to be
less than 90◦ , we see that the x-coordiante xB of B must be
p
2
2
2
xB = cos α x0 + y0 tan α − (ρ/ cos α) − (y0 − x0 tan α)
Since B lies on m4 , its is then easy to determined the y-coordinate yB of B:
p
2
2
yB = xB tan α = sin α cos α x0 + y0 tan α − (ρ/ cos α) − (y0 − x0 tan α)
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STANISLAV JABUKA
1.3. Step 3 with α 6= 90◦ and α + β 6= 90◦ . We continue with the assumption that α < 90◦
and for the moment also suppose that α + β 6= 90◦ . Note that the slope of the Euclidean
line passing through B and X is given by
yB − y0
xB − x0
so that the slope of the tangent line m5 to δ at B has slope
−
xB − x0
yB − y0
If yB = y0 then B and X are at the same height and so m5 is a vertical line. This can only
happen if α + β = 90◦ , a case that is currently excluded from consideration. On the other
hand, the slope of m5 must also equal tan(α + β) leading to the following master equation:
tan(α + β) = −
xB − x0
yB − y0
Unfortunately this equation is too difficult to solve for b in general.
1.4. Step 3 with α 6= 90◦ and α + β = 90◦ . In this case yB = y0 becomes the master
equation:
p
y0 = sin α cos α x0 + y0 tan α − (ρ/ cos α)2 − (y0 − x0 tan α)2
an equation that again it too cumbersome to solve in its generality. However, several special
cases are quite accessible, one of which we address in the following example.
1.4.1. Example: α = β = 45◦ . In this case the equation y0 =
√yB becomes quite manageable.
Namely, since in this case tan α = 1 and cos α = sin α = 1/ 2, the formula for yB becomes
p
1
yB =
x0 + y0 − 2ρ2 − (y0 − x0 )2
2
making the equation y0 = yB simplify as
p
1
y0 =
x0 + y0 − 2ρ2 − (y0 − x0 )2
2
p
2y0 = x0 + y0 − 2ρ2 − (y0 − x0 )2
p
y0 − x0 = − 2ρ2 − (y0 − x0 )2
(y0 − x0 )2 = 2ρ2 − (y0 − x0 )2
(y0 − x0 )2 = ρ2
y0 − x0 = ±ρ
Since we’ve assumed that α = 45◦ we have that y0 < x0 so that the above equation actually
becomes y0 − x0 = −ρ. Replacing x0 , y0 , ρ with their expressions in β and b we get the
HYPERBOLIC TRIGONOMETRY
5
equations:
r 2 − b2
r 2 + b2
r 2 − b2
−
=−
/ · 2b
2b tan γ
2b
2b sin γ
1
1
1
1
2
2
−b 1 +
+
=r 1−
−
tan γ sin γ
tan γ sin γ
sin γ + tan γ − sin γ tan γ
sin γ + tan γ + sin γ tan γ
b=r
Thus, as expected, d(AC) is independent of r and only depends on γ.
1.5. Step 2 with α = 90◦ . In the case of α = 90o the point B lies on the intersection of
the y-axis and the circle δ. Thus, to find the y-coordinate yB of B, we plug xB = 0 into
equation (1) and solve for y:
s
2 2
2
2
2
r −b
r 2 − b2
r + b2
y=
±
−
2b tan γ
2b sin γ
2b
The solution y which yields yB is gotten by choosing the minus sign, i.e.
s
2 2
2
2
2
r 2 − b2
r + b2
r −b
−
−
yB =
2b tan γ
2b sin γ
2b
Note that this can equivalently be written as
s
yB = y0 −
r 2 − b2
2b sin γ
2
− x20
The equation of the line through C and X is determined by
line through B and X
:
y=
y0 − yB
x + yB
x0
and so the slope of the tangent line m5 to δ at B is
Slope of m5 = −
x0
y0 − yB
On the other hand, the slope of m5 is also given as
Slope of m5 = tan(α + β) = tan(90◦ + β) = −
Combining the two we get the master equation
tan β =
y0 − yB
x0
1
tan β
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STANISLAV JABUKA
To solve this equation we utilize the formulae for y0 , x0 , yB derived above. Specifically we
first work out the right-hand side of the above equation:
r
v
2
u 2 2 2
s
r2 −b2
2
r −b
2
− x0 u
u
2b
sin
γ
2b sin γ
r 2 − b2
y0 − yB
t
−1
=
=
−1=
r2 +b2 2
x0
x0
(r2 + b2 ) sin γ
2b
Putting this back into the equation tan γ = (y0 − yB )/x0 yields
s
2
r 2 − b2
tan β =
−1
(r2 + b2 ) sin γ
which we now proceed to solve for b:
2
r 2 − b2
1 + tan β =
(r2 + b2 ) sin γ
2
2 2
r − b2
sin γ
=
cos β
r 2 + b2
cos β − sin γ
b2 = r 2
cos β + sin γ
s
cos β − sin γ
b=r
cos β + sin γ
2
Given this value of b in terms of r, the hyperbolic length b = d(AC) is then given by
√
√
cos β + sin γ + cos β − sin γ
√
(2)
b = d(AC) = log √
cos β + sin γ − cos β − sin γ
Note that again d(AC) is independent of the radius r of Γ, rather, it only depends on β and γ.
We also remark that the term cos β − sin γ is by necessity positive whenever 0 < β + γ < 90◦ .
To see this, note that if we had cos β < sin γ, then we would also obtain cos γ < sin β:
/2
cos β < sin γ
cos2 β < sin2 γ
2
/ · (−1)
2
− cos β > − sin γ
2
/+1
2
1 − cos β > 1 − sin γ
√
sin2 β > cos2 γ /
sin β > cos γ
Using these two inequalities, we’d arrive at a contradiction as
cos(β + γ) = cos β cos γ − sin β sin γ
< sin γ sin β − sin β sin γ
=0
HYPERBOLIC TRIGONOMETRY
7
Since the cosine function is non-negative on [0, 90◦ ], the above inequality cos(β + γ) <
0 implies that β + γ > 90◦ , a contradiction. Thus we are automatically ensured that
cos γ − sin β ≥ 0.
We proceed by putting equation (2) into a more manageable form. Namely:
√
√
cos β + sin γ + cos β − sin γ
√
b = log √
cos β + sin γ − cos β − sin γ
√
√
√
√
cos β + sin γ + cos β − sin γ
cos β + sin γ + cos β − sin γ
b
√
√
e =√
·√
cos β + sin γ − cos β − sin γ
cos β + sin γ + cos β − sin γ
p
(cos β + sin γ) + 2 cos2 β − sin2 γ + (cos β − sin γ)
=
(cos β + sin γ) − (cos β − sin γ)
p
cos β + cos2 β − sin2 γ
=
sin γ
With this equation in hand, we proceed to compute the hyperbolic cosine cosh b:
!
p
2 β − sin2 γ
1 b
cos
1
sin
γ
cos
β
+
p
cosh b = (e + e−b ) =
+
2
2
sin γ
cos β + cos2 β − sin2 γ
p
cos2 β + 2 cos β cos2 β − sin2 γ + cos2 β − sin2 γ + sin2 γ
=
p
2 sin γ cos β + cos2 β − sin2 γ
p
2 cos β cos β + cos2 β − sin2 γ
=
p
2 sin γ cos β + cos2 β − sin2 γ
=
cos β
sin γ
Thus we have shown:
Proposition 1.1. In a hyperbolic triangle ∆ABC with a right angle at A, the hyperbolic
lengths b = d(AC) and c = d(AB) can be computed in terms of the angle measures β = (∠B)◦
and γ = (∠C)◦ as
cos β
cos γ
cosh b =
and
cosh c =
sin γ
sin β
2. Second approach - Using the Beltrami-Klein model
In this section we approach the same goal, that of finding the hyperbolic lengths d(AB),
d(AC) and d(BC) of the edges of a hyperbolic triangle ∆ABC in terms of their angle
measures α, β, γ, but we do so in an entirely different manner. For the moment we restrict
our attention to right triangles but make slightly different arrangement than in the previous
section. Namely, we shall assume, without loss of generality, that A is again the center of Γ
(where Γ is still the circle for he Poincaré hyperbolic disk model), that C again lies on the
positive portion of the x-axis but that this time ∠C is a right angle (i.e. we take γ = 90◦ ),
see figure 2.
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STANISLAV JABUKA
Γ
B
A
δ
C
Figure 2. The triangle ∆ABC with its right angle at the vertex C.
We shall also rely on the abbreviations (some of which have already been introduced
above):
a = BC
a = d(BC)
b = AC
b = d(AC)
c = AB
c = d(AC)
Remember also that
b = d(AC) = ln
r+b
r−b
=⇒
eb =
r+b
r−b
From this, an easy check reveals that
eb − e−b
1 r+b r−b
2rb
=
−
= 2
sinh b =
2
2 r−b r+b
r − b2
eb + e−b
1 r+b r−b
r 2 + b2
cosh b =
=
+
= 2
2
2 r−b r+b
r − b2
sinh eb
2rb
tanh b =
=
cosh eb
r 2 + b2
Identical equations (with the proper replacements of symbols) relate c to c and a to a.
HYPERBOLIC TRIGONOMETRY
9
We will choose to use Γ also as the circle for the Beltrami-Klein model. Let D be the
interior of Γ and let F : D → D be the isomorphism from the Poincaré model to the
Beltrami-Klein model described in class, scaled so that the circles in both models coincide.
It is then not hard to verify that
F (x, y) =
2r2
(x, y)
r 2 + x2 + y 2
Let X 0 = F (X) for X = A, B, C. Notice that A = A0 since it coincides with the center of Γ,
see figure 3.
Γ
F
Γ
B0
B
A
A0
C
The Poincaré disk model.
C0
The Beltrami-Klein model.
Figure 3. The triangles ∆ABC and ∆A0 B 0 C 0 in the Poincaré and BeltramiKlein models.
Because F preserves right angles, we have ∠ACB ∼
= ∠A0 C 0 B 0 , making ∆A0 B 0 C 0 a right
(Euclidean) triangle with a right angle at C. Moreover, since diamaters in Γ get mapped to
the same diameters in Γ and since AB lies on such a diameter in Γ, we conclude that A0 B 0
also lies on that same diameter. This in turn shows that the angles ∠BAC and ∠B 0 A0 C 0
are congruent, i.e. (∠B 0 A0 C 0 )◦ = α. In particular we have
cos α =
A0 C 0
A0 B 0
Since C = (b, 0) then
2rb
2r2
(b, 0) =⇒ A0 C 0 = r 2
= r tanh b
C = F (C) = 2
2
r +b
r + b2
In the same way one finds that
A0 B 0 = r tanh c
Putting the above together we arrive at
tanh b
cos α =
tanh c
0
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STANISLAV JABUKA
Let B 00 be the inverse of B under inversion in Γ and let δ be the circle perpendicular to Γ
and passing through B and C, see figure 4.
B 00
Γ
B1
β
B
β
A
δ
O1
C
C 00
Figure 4. The triangles ∆ABC and various points derived from it: The
points B 00 and C 00 are the inversed of B and C under inversion in Γ. O1 is the
center of δ, the unique circle perpendicular to Γ and containing B and C (and
thus also B 00 and C 00 ). Finally, B1 is the midpoint of the Euclidean segment
BB 00 . The three right angles in the picture are indicated by a little square.
Since δ ⊥ Γ it must be that B 00 lies on δ. Moreover
BB 00 = AB 00 − AB =
r2
r 2 − c2
2r
− AB =
=
c
sinh c
AB
and likewise
CC 00 =
2r
sinh b
Let B1 be the midpoint of the Euclidean segment BB 00 so that O1 B1 is perpendicular to
BB 00 (where O1 is the center of δ). Since O1 B is also perpendicular to the tangent line to δ
at B, we obtain
β = (∠B)o = 90o − (∠B1 BO1 )o = (∠BO1 B1 )
HYPERBOLIC TRIGONOMETRY
11
and so (using CC 00 = 2O1 C since CC 00 is a diameter of δ, because of γ = 90◦ )
sin β =
sinh b
BB1
BB 00
BB 00
=
=
=
00
sinh c
BO1
2CO1
CC
To summarize:
Theorem 2.1. Let ∆ABC be a right triangle in hyperbolic geometry with the right angle at
C. Then
sinh a
tanh b
and
sin(α) =
cos(α) =
tanh c
sinh c
tanh a
sinh b
cos(β) =
and
sin(β) =
tanh c
sinh c
Using these formulae and hyperbolic trig identities one can derive additional useful relations:
sinh2 a tanh2 b
1 = sin2 α + cos2 α =⇒ 1 =
+
sinh2 c
tanh2 c
Solving for sinh2 c gives
cosh2 c = 1 + sinh2 c = 1 + sinh2 a + cosh2 c · tanh2 b = cosh2 a + cosh2 c · tanh2 b
Multiplying by cosh2 b yields
cosh2 b cosh2 c = cosh2 a · cosh2 b + cosh2 c · sinh2 b
cosh2 c (cosh2 b − sinh2 b) = cosh2 a · cosh2 b
cosh2 c = cosh2 a · cosh2 b
cosh c = cosh a · cosh b
In addition
cos α
tanh b sinh c
cosh c
=
·
=
= cosh a
sin β
tanh c sinh b
cosh b
We have then proved the following statements:
Corollary 2.2. Let ∆ABC be a right triangle in hyperbolic geometry with the right angle
at C. Then
cos(∠A)
cos(∠B)
and
cosh d(AC) =
cosh d(BC) =
sin(∠B)
sin(∠A)
cos(∠A) cos(∠B)
cosh d(AB) = cosh d(BC) · cosh d(AC) =
·
= cot(∠A) · cot(∠B)
sin(∠B) sin(∠A)
This corollary explains how to calculate the Poincaré lengths of the edges of a right triangle
in terms of its angles. These formulae do not have Euclidean analogues.
Note that the results from corollary 2.2 agree with those found in proposition 1.1 (keeping
in mind that ∠A was the right angle in the setting of proposition 1.1).
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STANISLAV JABUKA
Theorem 2.3. Let
(∠A)◦ , β = (∠B)◦
relations:
sin β
sin α
=
=
sinh a
sinh b
∆ABC be an arbitrary hyperbolic triangle with angle measures α =
and γ = (∠C)◦ . Then the hyperbolic lengths of its sides satisfy the
sin γ
sinh c
cosh c = cosh a cosh b − sinh a sinh b cos γ
cosh c =
The hyperbolic law of sine.
The hyperbolic law of cosine.
cos α cos β + cos γ
sin α sin β
While the first two of these relations have obvious analogues in Euclidean geometry, the
third one does not have a Euclidean counterpart.