Homework assignment 1 Differential Geometry I, Fall 2013

HOMEWORK ASSIGNMENT 1

1. Bookkeeping questions

Please remember that the material of the bookkeeping questions is examinable, except for the challenge
questions and those that explicitly say otherwise. There will be written solutions for all the bookkeeping
questions that are examinable material.

Exercise 1.1. [Rigid motions] A rigid motion of Euclidean space is a function R : Rn → Rn such that
|R(x) − R(y)| = |x − y| for all x, y ∈ Rn .

It is well-known from linear algebra that an affine rigid motion is a composition of a translation, a rotation,
and possibly a reflection in a hyperplane. (Here we agree that rotations are orientation preserving, i.e.
elements of the special orthonormal group.)

Show that every rigid motion is affine. Hint: The composition of two rigid motions of Rn is again a rigid
motion. Consider a rigid motion R : Rn → Rn with R(0) = 0. Explain why {R(e1 ), . . . , R(en )} is an
orthonormal basis of Rn . Think about the extent to which a point x ∈ Rn is determined by the distances
|x − e1 |, . . . , |x − en |?

Solution: Let R : Rn → Rn be a rigid motion. We want to show that R is affine. By composing R with the
translation by −R(0) we can reduce to the case where R(0) = 0. It remains to show that such a rigid motion
is linear. If we insert y = 0 in the defining equation of rigid motions, we get |R(x)| = |x| or equivalently that
R(x) · R(x) = x · x for all x ∈ Rn . If, on the other hand, we square the defining equation we obtain that
R(x) · R(x) − 2R(x) · R(y) + R(y) · R(y) = x · x − 2x · y + y · y.
Thus R(x) · R(y) = x · y for all x, y ∈ Rn . It follows that R maps the orthonormal basis e1 , . . . , en to another
orthonormal basis R(e1 ), . . . , R(en ). It follows that for all x = (x1 , . . . , xn ) ∈ Rn we have that
Xn Xn Xn
R(x) = (R(x) · R(ei )) R(ei ) = (x · ei ) R(ei ) = xi R(ei ).
i=1 i=1 i=1
Thus R is linear, as required.

Exercise 1.2. [Cycloid] Let a ∈ [0, 1]. A marker is attached to a spike of a wheel with unit radius at
distance a from the axis. The wheel is rolling with constant angular velocity −1 along the x-axes from −∞
to +∞ so that the marker is at (0, 1 − a) when the wheel touches the origin. Describe the trajectory of the
marker.

Solution: We compute the motion of the marker from the motion of the center of the wheel and the marker’s
position relative to this center. The center of the wheel moves on the line {(x, 1) : x ∈ R} with constant
velocity to the right. As the wheel has radius r = 1 and angular velocity ω = −1, the point on the wheel
where it touches the ground has velocity v = rω = −1 in x-direction. As there is no sliding, the center of the
wheel moves with constant velocity 1. If at time t = 0 the wheel touches the origin, then the center of the
week is at the point c(t) = (t, 1) at time t. Relative to the center of the wheel, the marker moves at constant

Date: October 9, 2013.

Please do not hesitate to consult your assistant if you need help

1
Homework assignment 1 Differential Geometry I, Fall 2013

radius a and constant angular velocity −1. Hence its relative position at time t is r(t) = (−a sin(t), −a cos(t))
and the trajectory of the marker is parametrized by m(t) = c(t) + r(t) = (t − a sin(t), 1 − a cos(t)).

Exercise 1.3. [Viviani’s window] Give an explicit parametrization for a regular curve whose trace is the
intersection of {(x, y, z) ∈ R3 : (x − 1)2 + y 2 = 1} and {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 4}. Write down an
integral that computes its length. Recommendation: Try to sketch this curve by hand.

Solution: Putting the two equations together we find

z 2 = 4 − 2x .

If we parametrize the curve we are tempted to use coordinates adapted to the cylinder, i.e. we would like
to write x(t) = 1 + cos(t) and y(t) = sin(t). The above condition tells us that one part of the trace of
the curve lies above the xy-plane, the other below. Both parts of the trace of the curve are joined when
z = 0, i.e. x = 2. To understand what is happening at that point we linearize x = 2 − ε. Inserting into the
cylinder condition yields y 2 = 2ε up to order of ε. Inserting into the condition for the sphere, we end up
with −y 2 + z 2 = 0. So, around the point (2, 0, 0) the trace of the curve looks like two intersecting lines. One
is leading from the southern hemisphere into the northern one. The other line goes from north to south.
But above we saw that we have two parts of the curve that are connected at the point (2, 0, 0). Let us try to
use the parametrization for x and y as above. Then, z 2 = 4 − 2(1 + cos(t)) = 2(1 − cos(t)).
p As we start at
t = 0 we move into the northern hemisphere. So we select the positive root for z = p 2(1 − cos(t)). As we
reach t = 2π we want to move into the southern hemisphere. Then we select z = − 2(1 − cos(t)). As we
reach 4π we have traversed the trace of the curve completely. Now we note that cos(2θ) = 1 − 2 sin2 (θ) [2].
Thus z(t) = 2 sin(t/2) automatically selects the correct hemisphere. Therefore, one valid parametrization is

c(t) = (1 + cos(t), sin(t), 2 sin(t/2)) .

The velocity is ċ(t) = (− sin(t), cos(t), cos(t/2))γ and the length

Z 4π p
length(c) = 1 + cos2 (t/2) dt .
0

Figure 1 shows what the trace of the curve looks like.

Figure 1. Sections of the trace of Viviani’s curve [1]

Exercise 1.4. [Cauchy-Crofton formula] Let c : I → R2 be a planar curve. For every θ ∈ R let vθ =
(cos θ, sin θ) and cθ : I → R2 be the curve t 7→ (c(t) · vθ ) vθ . Geometrically, what is cθ ? Assume that I is
compact. Show that
Z 2π
length(cθ ) dθ = 4 length(c).
0

Please do not hesitate to consult your assistant if you need help

2
Homework assignment 1 Differential Geometry I, Fall 2013

Solution: cθ is the curve we get by composing c with the orthogonal projection onto the line spanned by
vθ . Note that ċθ (t) = (ċ(t) · vθ )vθ , |ċθ (t)| = |ċ(t) · vθ |. Let I = [a, b]. Then we compute
Z 2π Z 2π Z b Z b Z 2π
length(cθ )dθ = |ċθ (t)|dtdθ = |ċ(t) · vθ |dθdt,
0 0 a a 0

using Fubini and our formula for |ċθ (t)|. Hence it suffices to show that for every vector v ∈ R2 we have
Z 2π
|v · vθ |dθ = 4|v|
0

and to apply this to v = ċ(t). This we leave to the reader. Hint: Reduce to the case v = (0, r) for r > 0.

Exercise 1.5. Let 0 < a < b. Compute the evolute of the ellipse c : [0, 2π) → R2 given by c(θ) =
(a cos θ, b sin θ).

Solution: We have that

ċ = (−a sin θ, b cos θ)
c̈ = (−a cos θ, −b sin θ)
p
|ċ| = a2 sin2 θ + b2 cos2 θ =: v
−1
T = |ċ| ċ = v −1 (−a sin θ, b cos θ)
N = iT = v −1 (−b cos θ, −a sin θ)
−3
κ = |ċ| (c̈ · iċ) = v −3 ab
 2
a − v2 b2 − v 2

e = c + κ−1 N = cos θ, sin θ .
a b

Exercise 1.6. [Frénet curves] Let c : I → Rn be a curve.

For t ∈ I and i = 1, . . . , n − 1 we let

E i (t) = spanR {c0 (t), c00 (t), . . . , c(i) (t)}.
Also let E 0 (t) = {0} ⊂ Rn and E n (t) = Rn .

Show that if ψ : J → I is a diffeomorphisms of intervals, then Eci ◦ ψ = Ec◦ψ

i
for all i = 0, 1, . . . , n.

The curve c : I → Rn is said to be Frénet if E n−1 (t) has dimension n − 1 for all t ∈ I. The Frénet frame of
a Frénet curve is the family of vector fields e1 , . . . , en : I → Rn that is uniquely determined by the following
conditions to hold for every t ∈ I: the collection {e1 (t), . . . , en (t)} is a positively oriented orthonormal basis
of Rn and for every i = 1, . . . , n − 1 we have that spanR {e1 (t), . . . , ei (t)} = E i (t) and ei (t) · c(i) (t) > 0.

For convenience, we let e0 , en+1 : I → Rn be the trivial vector fields with constant value zero.

(i) Explain why there exists a unique Frénet frame. How does it change with orientation preserving
reparametrizations of the curve?
(ii) Explain why e0i · ej = −e0j · ei for all i, j = 1, . . . , n.
(iii) Explain why e0i (t) ∈ E i+1 (t) for all t ∈ I and all i = 1, . . . , n − 1.

From now on, we assume that c : I → Rn is a unit speed Frénet curve. Let κ1 , . . . , κn−1 : I → R be the
smooth functions κi = e0i · ei+1 . For convenience, we let κ0 = κn = 0. Explain why
e0i = −κi−1 ei−1 + κi ei+1 for all i = 1, . . . , n.
These are the (general) Frénet formulae.

Please do not hesitate to consult your assistant if you need help

3
Homework assignment 1 Differential Geometry I, Fall 2013

Show that the trace of a Frénet curve lies in an affine hyperplane if and only if κn−1 = 0.

Remark and advice: Let c : I → R3 be a unit speed Frénet curve. Then e1 = T = c0 is called the tangent
00
field, e2 = N = |cc00 | the normal field, and e3 = B = e1 × e2 the binormal field of the curve. Moreover, κ1 = κ
is called the curvature and κ2 = τ is called the torsion of the curve. Try to get a feel for the Frénet formulae
in this special case first.

Solution: If ψ : J → I is a diffeomorphism of intervals, we show inductively: For all i = 0, . . . , n − 1 there

Pi
exist smooth functions αij : J → R, such that (c ◦ ψ)(i) (s) = j=0 αij (s)c(j) (ψ(s)) and αii = 1. For i = 0
choose α00 = 1. Assume the statement is true for some i, then
i
X
(c ◦ ψ)(i+1) = (αij )0 c(j) ◦ ψ + αij ψ 0 c(j+1) ◦ ψ,
j=0

which after rearranging has the desired form with the coefficient function of c(i+1) (ψ(s)) being αii = 1 by
induction. This shows Ec◦ψ i
⊂ Eci ◦ ψ. For the other inclusion we use that ψ −1 is also a diffeomorphism and
i i
we get Ec ◦ ψ = E(c◦ψ)◦ψ−1 ◦ ψ ⊂ Ec◦ψ i
◦ ψ −1 ◦ ψ = Ec◦ψ as desired.
n
Assume that the curve c : I → R is Frénet. Note that c is Frénet if and only if for all t ∈ I we have
c0 (t), c00 (t), . . . , c(n−1) (t) are linearly independent. In particular, c(i) (t) ∈
/ E i−1 (t) for i = 1, . . . , n − 1.

(i) We construct vector fields e1 , . . . , en : I → Rn which satisfy the conditions of a Frénet frame:
0
Choose e1 = |cc0 | . It has unit length and satisfies e1 · c0 > 0. Assume now we have already chosen
e1 , . . . , ei−1 , then we define
i−1 
(i)
X  ẽi
ẽi = c − c(i) · ej ej , ei = .
j=1
|ẽi |

(Note: This is exactly Gram-Schmidt orthonormalization!) Then we see immediately that ei is a

smooth vector field, that e1 (t), . . . , ei (t) is an orthonormal basis of E i (t), and that
 
i−1
(i) 1  (i) 2 X (i) 2
ei · c = |c | − (c · ej ) > 0,
|ẽi | j=1

by the theorem of Pythagoras (using that e1 (t), . . . , ei−1 (t) can be enriched to an orthonormal basis and
c(i) (t) ∈
/ E i−1 (t) because c is Frénet). After choosing e1 , . . . , en−1 as indicated above, for every t ∈ I
there is exactly one choice of vector en (t) such that e1 (t), . . . , en (t) is a positively oriented orthonormal
basis, and it depends smoothly on t (again Gram-Schmidt). We have checked in every step that our
constructed frame has the desired properties. Conversely note that in every stage of the construction
we had exactly two possibilities to choose ei (t) such that e1 (t), . . . , ei (t) are an orthonormal basis of
E i (t) and the condition ei · c(i) > 0 makes the choice unique.
(ii) Note that ei ·ej = δij is constant for all i, j. Thus taking the derivative we get: 0 = (ei ·ej )0 = e0i ·ej +ei ·e0j .
(iii) From the construction we see that ei is a C ∞ (I)-linear combination of c0 , . . . , c(i) , i. e. there exist
smooth functions α1 , . . . , αi on I such that ei = α1 c0 + . . . + αi c(i) . Using the product rule, we see
e0i (t) ∈ spanR {c0 (t), . . . , c(i+1) (t)} = E i+1 (t).

Pi+1
For the general Frénet formulae we note that e0i (t) ∈ E i+1 (t), thus e0i = j=1 (e0i · ej ) ej , as the ej (t) form
an orthonormal basis. Thus it suffices to check the following four identities, using (ii) in the first three
deductions:

• e0i · ej = −e0j · ei = 0 for j < i − 1. This follows as e0j (t) ∈ E j+1 (t) and ei (t) ⊥ E j+1 (t)
• e0i · ei−1 = −e0i−1 · ei = −κi−1 by definition
• e0i · ei = −e0i · ei , thus e0i · ei = 0
• e0i · ei+1 = κi by definition.

Please do not hesitate to consult your assistant if you need help

4
Homework assignment 1 Differential Geometry I, Fall 2013

Finally note that the trace of c lies in an affine hyperplane if and only if there exists a unit vector v ∈ Rn ,
λ ∈ R such that c(t) · v = λ for all t ∈ I. Taking the derivatives of the equation we see this is equivalent
to the existence of a unit vector v such that c(i) (t) · v = 0 for i = 1, . . . , n − 1, that is v ⊥ E n−1 (t) for all
t ∈ I. On the other hand we have e0n = −κn−1 ei−1 . Thus κn−1 = 0 is equivalent to en being constant. If
this is the case, we can choose v = en . On the other hand, if we have v as above, we may choose en = v or
en = −v depending on the orientation. This proves the equivalence.

Exercise 1.7. [Osculating circle and circumcircles] Let c : I → R2 be a regular curve and let t0 ∈ I be an
interior point with κ(t0 ) 6= 0. Show that the osculating circle of c at t0 is the limit of the circumcircles of
the triangles with vertices c(t0 ), c(t1 ), c(t2 ) as t1 , t2 → t0 where t0 , t1 , t2 ∈ I all distinct.

Hint: Reduce to the case where the curve has unit speed and where t0 = 0 and c(t0 ) = 0. Show that the
circumcenter of a triangle with vertices x, y, z ∈ R2 is given by
x+z (y − z) · (y − x) i(x − z)
(1) +
2 (y − z) · i(x − z) 2
where i : R2 → R2 is counterclockwise rotation by π/2. The following analysis fact may also be useful. If
f : I → R is smooth and f (0) = 0, then the limit
f (t2 )/t2 − f (t1 )/t1
lim
t1 ,t2 →0 t2 − t1
0,t1 ,t2 distinct

exists. Why? What is the limit equal to?

Solution: Let m be the point given by the formula (1). To show that m is the circumcenter of the triangle
with vertices x, y, z you can calculate the distances of m to x, y, z and show they are equal. Alternatively
you can also use that the circumcenter is the intersection of the perpendicular bisectors of the sides of the
triangle and show that (m − x+y x+z
2 ) ⊥ (x − y), (m − 2 ) ⊥ (x − z). One way of direct construction of this
x+y x−y
point is as follows: We are looking for α, β ∈ R such that x+z x−z
2 + αi 2 = 2 + βi 2 . Taking the inner
product of both equations with x − y eliminates β. Solving for α and inserting gives exactly the expression
in (1).
For the analysis fact: If f : I → R is smooth with 0 ∈ I, f (0) = 0, then by Hadamard’s lemma there is
g : I → R smooth such that f (t) = tg(t). Hence t 7→ f (t)/t extends to the smooth function g. For the
derivative at 0 we take the second derivative of the equation f (t) = tg(t) and get f 00 (t) = 2g 0 (t) + tg 00 (t). So
at t = 0 we obtain g 0 (0) = 12 f 00 (0). The existence of the limit follows from the mean value theorem and it is
equal to g 0 (0).
Now we turn to proving that the osculating circle at t0 ∈ J is the limit of the circumcircles of the triangles
with vertices c(t0 ), c(t1 ), c(t2 ) as t1 , t2 → t0 in I. To see this, we assume that t0 = 0, c unit speed and
c(t0 ) = 0 by reparametrization of c and composition with a translation. The function s 7→ c(s)/s is smooth
across the origin with derivative 12 c̈(0) = κ(0) 2 iċ(0). Here we used the first Frénet formula. It follows that
c(t2 ) c(t2 )−c(t1 )
c(t1 ) c(t2 )(c(t2 ) − c(t1 )) ic(t1 ) t2 t2 −t1 ic(t1 )
lim + = lim c(t2 ) c(t1 )
t1 ,t2 →0 2 ic(t1 ) · c(t2 ) 2 t1 ,t2 →0 t2 − t1 2t1
t1 6=t2 | {z } t1 6=t2 i c(tt11 ) · t2 −t1
Circumcenter of the triangle
with vertices 0=c(0), c(t1 ), c(t2 )

ċ(0) · ċ(0) ċ(0) 1

= i = iċ(0).
(iċ(0)) · ( κ(0)
2 iċ(0))
2 κ(0)

But this is exactly the center of the osculating circle at 0.

Exercise 1.8. [Height function] Let c : I → R2 be a regular curve and let v ∈ R2 be a unit vector. The
height function h : I → R is given by t 7→ v · c(t). Show that t ∈ I is a critical point of h if and only if the
tangent of c at t is perpendicular to v. Discuss how the nature of such a critical point is related to the sign
of the geodesic curvature. If, on the other hand, the height function h : I → J = h(I) has no critical points,

Please do not hesitate to consult your assistant if you need help

5
Homework assignment 1 Differential Geometry I, Fall 2013

it has a smooth inverse ψ : J → I by the inverse function theorem. The reparametrization c ◦ ψ : J → R2 is

special. In descriptive, non-technical language, how?

d
Solution: A point t0 ∈ I is critical for h and and only if 0 = dt |t=t0 v · c(t) = v · ċ(t0 ) if and only if the
tangent of c at t0 is perpendicular to v. To analyse the nature of the critical point, we may assume that ċ(t0 )
and v form an oriented orthonormal basis, i.e. v = iċ(t0 ), where i is the rotation of the plane by π2 . Then
the second derivative of h at t0 is given by v · c̈(t0 ) = (iċ(t0 )) · c̈(t0 ) = κ(t0 )|ċ(t0 )|3 . Thus if the geodesic
curvature is positive, the critical point is a minimum and if the curvature is negative, it is a maximum.
If we compute the new heigth function h with respect to the reparametrization c◦ψ, we see h(s) = v·c(ψ(s)) =
h(ψ(s)) = s. Thus the new parametrization describes the curve as a graph above the line spanned by v.

Exercise 1.9. [Degree in dimension 1]

Let S 1 = {x ∈ R2 : |x| = 1}. Let p : R → S 1 be the function θ 7→ (cos θ, sin θ).

Let f : R → S 1 be a continuous function.

The following fact is as fundamental as it is elementary. If you have not thought about it before, now is a
good moment.

There exists a unique continuous function fˆ : R → R such that p ◦ fˆ = f . A function fˆ with this property
is called a lift of f . Here is a useful picture.
R
-

fˆ
p

?
f
R - S1
The lift of f is uniquely determined once we agree on its value at a point. In particular, the lift of a
continuous function R → S 1 is unique modulo integer multiples of 2π.

Let fˆ : R → R be a lift of f .

If f is continuously differentiable (smooth), then so is its lift fˆ. Moreover,

−ydx + xdy
Z
(2) fˆ(b) − fˆ(a) =
f |[a,b] x2 + y 2

for all a, b ∈ R with a < b.

How is this related to the definition of the winding number of a smooth closed curve that avoids the origin
in complex analysis?

Assume now that f is 2π periodic.

It is natural to think of f as a continuous function S 1 → S 1 . We do not push this point of view here.

There exists k ∈ Z such that fˆ(2π) − fˆ(0) = 2πk. This integer k is independent of the choice of lift. It is
called the degree of the continuous periodic function f : R → S 1 . Note that fˆ(t + 2π) − fˆ(t) = 2πk for all
t ∈ R.

From your complex analysis class you remember the formula

Z
1 dz
2πi c z

Please do not hesitate to consult your assistant if you need help

6
Homework assignment 1 Differential Geometry I, Fall 2013

for the winding number of a continuously differentiable curve c : [a, b] → C \ {0}. Make the relation with (2)
above.

Let h : [0, 1] × R → S 1 continuous be such that the functions fs : R → S 1 given by t 7→ h(s, t) are 2π
periodic for all s ∈ [0, 1]. We say that h is a homotopy between the functions f0 , f1 : R → S 1 . We say that
continuous and 2π periodic functions R → S 1 are homotopic if there exists a homotopy between them.

Show that two continuous and 2π periodic functions f0 , f1 : R → S 1 are homotopic if and only if they have
the same degree. Hint: For one direction, use that [0, 1] × R is simply connected. For the other direction,
start by constructing a (very simple) homotopy of the lifts fˆ0 , fˆ1 .

The degree of a continuous and L periodic function g : R → R2 \{0} is defined as the degree of the continuous
and 2π periodic function
L


g 2π t
t 7→ L

.
g 2π t

Solution:

First let us establish the relation between the complex analysis winding number and the differential geometry
one.

Z Z
1 dz 1 dx + idy
= z
2πi c z 2πi c zz
−ydx + xdy
Z Z
1 xdx + ydy i
= +
2πi c x2 + y 2 2πi c x2 + y 2
−ydx + xdy
Z
1
=
2π c x2 + y 2
where the first term in the second line is zero by the fundamental theorem of line integrals, because we
integrate the exact form (xdx + ydy)/(x2 + y 2 ) = d ln(x2 + y 2 )/2 over a periodic curve.

Now we turn to proving the main result of this exercise.

Let h : [0, 1] × R → S1 be a homotopy from f0 to f1 . Then as [0, 1] × R is simply connected, there is a lift
ĥ : [0, 1] × R → R such that h = p ◦ ĥ. Then, as hs is 2π periodic, we have that ĥs (2π) − ĥs (0) ∈ 2πZ.
Now s 7→ ĥs (2π) − ĥs (0) is a continuous function from a connected space with values in a discrete space.
This implies that this function is constant. Hence, 2πks = ĥs (2π) − ĥs (0) = fˆ0 (2π) − fˆ0 (0) = 2πk0 and in
particular k0 = k1 . This proves the first part of the result.

For the converse, choose lifts fˆ0 , fˆ1 . As the degree of the two maps are the same, we have fˆ0 (2π) − fˆ0 (0) =
fˆ1 (2π) − fˆ1 (0). Now define ĥs : [0, 1] × R : (s, t) 7→ sfˆ0 (t) + (1 − s)fˆ1 (t) and notice that ĥs (2π) − ĥs (0) =
fˆ0 (2π) − fˆ0 (0) ∈ 2πZ. Thus, hs = p ◦ ĥs is a 2π-periodic function (Why does this not work if the degrees
are different?) and h = p ◦ ĥ homotopes f0 to f1 .

2. Project

Our first project this semester is dedicated to the study of doodles. In 2011, I organized a class at MIT
where, each time, a different faculty member gave an introduction to their area of research. Most students
were in the first year of their undergraduate study. The idea was to connect their introductory classes on

Please do not hesitate to consult your assistant if you need help

7
Homework assignment 1 Differential Geometry I, Fall 2013

linear algebra and analysis (“calculus”) with what is out there. I really liked Haynes Miller’s presentation
on the Whitney-Graustein theorem. His lecture notes are a model of mathematical exposition. In this first
project, I’m asking you to work through (part) of his notes. In some places, Haynes’ notation is different
from ours. For example, he calls a homotopy between doodles through doodles an isotopy, “isos” meaning
“the same” or “equal” in Greek. Haynes has kindly given me permission to share his materials with you. I’m
asking you to read carefully through the notes and to supply a rigorous proof for all his claims. Translate
everything into our language on the way. I hope you will be as playful with this project as we were at MIT!
In the exercise class, you’ll go through all these notions together. Ideally, it will be a discussion moderated
by the assistants, and not one led by them. As every project, this is examinable material.

You will find Haynes’ lecture notes with slight annotations here: https://www.math.ethz.ch/education/
bachelor/lectures/hs2013/math/diff_geometrie/doodlesannotated.pdf.

References

1. Eric W. Weisstein, Viviani’s curve. from mathworld–a wolfram web resource, 2012, [Online; accessed 24-September-2012].
2. Wikipedia, List of trigonometric identities — Wikipedia, the free encyclopedia, 2012, [Online; accessed 24-September-2012].

Please do not hesitate to consult your assistant if you need help