Similar Polygons

In today's blog, I present a proof from Euclid on similar polygons. I use this theorem later in my proof for the existence of pi.

Definition: Similar Polygons

Two polygons are similar if corresponding angles are congruent and corresponding sides are proportional.

Lemma 1: A/B = C/D, then (A+C)/(B+D) = A/B = C/D

Proof:

(1) Let A/B = C/D

(2) Then AD=BC and A = BC/D

(3) So,

(A+C)/(B+D) = (BC/D + C)/(B + D) = (BC/D + CD/D)/(B+D) = [(BC+CD)/D][1/(B+D)] =
= [C(B+D)]/[D(B+D)] =C/D

QED

Corollary 1.1: A/B = C/D = E/F → (A+C+E)/(B+D+F)=A/B=C/D=E/F

Proof:

(1) (A+C)/(B+D) = A/B = C/D = E/F [See Lemma 1 above]

(2) Let U = A+C, V = B+D

(3) U/V = A/B = C/D [From step #1]

(4) U+E/V+F = A/B = C/D=E/F [From Lemma 1 above]

(5) But then:

(A + C + E)/(B + D + F)

QED

Lemma 2: if A/B = C/D, then A/C = B/D

Proof:

(1) A/B = C/D

(2) A*D = B*C

(3) Dividing both sides by C*D gives us:

A/C = B/D

QED

Lemma 3: if BE/AB = GL/GF and AB/BC = GF/GH,
then: BE/BC = GL/GH

Proof:

(1) BE/AB = GL/GF → BE*GF = GL*AB

(2) AB/BC = GF/GH → AB*GH = BC*GF

(3) So that:
BE*GF*AB*GH = GL*AB*BC*GF

(4) Dividing both sides by GF*AB*BC*GH gives us:
BE/BC = GL/GH

QED

Therem 1: Similar polygons have a ratio equal to the square of the ratio of two corresponding sides.




























Proof:

(1) Let ABCDE and FGHKL be similar polygons.

(2) From the properties of similar polygons (see definition above), we know that:

(a)
∠ ABC ≅ ∠ FGH
AB/BC = FG/GH

(b)
∠ BCD ≅ ∠ GHK
BC/CD = GH/HK

(c)
∠ CDE ≅ ∠ HKL
CD/DE = HK/KL

(d)
∠ DEA ≅ ∠ KLF
DE/EA = KL/LF

(e)
∠ EAB ≅ ∠ LFG
EA/AB = LF/FG

(3) From congruent angles and corresponding sides (see here), we know that:

(a) triangle ABC is equiangular with triangle FGH (from #2a)

(b) triangle BCD is equiangular with triangle GHK (from #2b)

(c) triangle ECD is equiangular with triangle LHK (from #2c)

(d) triangle ABE is equiangular with triangle FGL. (from #2e)

(4) BE/BC = GL/GH (from Lemma 3 above) since:

(a) BE/AB = GL/GF (from #3d)

(b) AB/BC = GF/GH (from #3a)

(5) ∠ EBC ≅ ∠ LGH since:

(a) ∠ ABE ≅ ∠ FGL (#3d)

(b) ∠ ABC ≅ ∠ FGH (#3a)

(c) ∠ EBC = ∠ ABC - ∠ ABE

(d) ∠ LGH = ∠ FGH - ∠ FGL = ∠ ABC - ∠ ABE

(6) triangle EBC ≅ triangle LGH (from here) since:

(a) BE/BC = GC/GH (#4)

(b) ∠ EBC ≅ ∠ LGH (#5)

(7) triangle BOC is equiangular with triangle GPH since:

(a) ∠ OBC ≅ ∠ PGH (from #3b)

(b) ∠ BCO ≅ ∠ GHP (from #6)

(8) From the properties equiangular triangles we know that:

(a) ∠ BAM ≅ ∠ GFN (#3a)

(b) ∠ ABM ≅ ∠ FGN (#3d)

(c) ∠ MBC ≅ ∠ NGH (#6)

(d) ∠ BCM ≅ ∠ GHN (#3a)

(e) ∠ ODC ≅ ∠ PKH (#3b)

(f) ∠ OCD ≅ ∠ PHK (#3c)

(9) From congruent angles and corresponding sides again (see here), we know that:

(a) triangle AMB equiangular with triangle FNG from ∠ BAM ≅ ∠ GFN (#7a) and ∠ ABM ≅ ∠ FGN (#7b)

(b) triangle BMC equiangular with triangle GNH from ∠ MBC ≅ ∠ NGH (#7c) and ∠ BCM ≅ ∠ GHN (#7d)

(c) triangle COD equiangular with triangle HPK from ∠ ODC ≅ ∠ PKH (#7e) and ∠ OCD ≅ ∠ PHK (#7f).

(9) From properties of equiangular triangles, we know that:

(a) CO/OD = HP/PK (from #8c)

(b) BO/OC = GP/PH (from #7)

(c) AM/MB = FN/NG (from #8a)

(d) BM/MC = GN/NH (from #8b)

(10) Using Lemma 3 above gives us:

(a) BO/OD = GP/PK (from #9b and #9a)

(b) AM/MC = FN/NH (from #9c and #9d)

(11) Since triangles with the same height have their areas proportional to their bases (see here), we know that:

(a) BO/OD = BOC/COD

(b) BO/OD = BOE/OED

(c) GP/PK = GPH/HPK

(d) GP/PK = LGP/LPK

(e) AM/MC = ABM/MBC

(f) AM/MC = AME/EMC

(g) FN/NH = FGN/NGH

(h) FN/NH = FNL/LNH

(12) From step #11, we find that:

(a) BOC/COD = BOE/EOD

(b) GPH/HPK = LGP/LPK

(c) ABM/MBC = AME/MEC

(d) FGN/NGH = FNL/LNH

(13) From Lemma 1 above, we get:

(a) ABM/MBC = ABE/CBE since [ABE/CBE = (ABM + AME)/(MBC + MEC)]

(b) FGN/NGH = FGL/HGL since [FGL/HGL = (FGN + FNL)/(NGH +LNH)]

(c) BOC/COD = CBE/CED since [CBE/CED = (BOC + BOE)/(COD + OED)]

(d) GPH/HPK = HGL/HLK since [HGL/HLK = (GPH + LGP)/(HPK + LPK)]

(14) Combining step #11 with step #13 gives us:

(a) BO/OD = CBE/CED (see #11a and #13c)

(b) GP/PK = HGL/HLK (see #11c and #13d)

(c) AM/MC = ABE/CBE (see #11e and #13a)

(d) FN/NH = FGL/HGL (see #11g and #13b)

(15) Combining step #10 with step #14 gives us:

(a) CBE/CED = HGL/HLK (see #14a, #14b and #10a)

(b) ABE/CBE = FGL/HGL (see #14c, #14d and #10b)

(16) Using step #15 and Lemma 2, we get:

(a) CBE/HGL = CED/HLK (from #15a)

(b) ABE/FGL = CBE/HGL (from #15b)

(c) Thus, CBE/HGL = CED/HLK = ABE/FGL

(17) Now, polygon ABCDE and polygon FGHKL divide up into three triangles where:

polygon ABCDE = triangle ABE + triangle CBE + triangle CED

polygon FGHKL = triangle FGL + triangle HGL + triangle HLK

(18) Using step #16, we can apply Corollary 1.1 above to get:

triangle ABE/triangle FGL = polygon ABCDE/polygon FGHKL

(19) Since similar triangles are one to another in square ratio of the corresponding sides (see here), we have (see step #3d):

triangle ABE/triangle FGL = (AB)²/(FG)²

(20) Combining step #18 with step #19 gives us:

polygon ABCDE/polygon FGHKL = (AB)²/(FG)²

QED

References

• David Joyce, Euclid's Elements, Book VI, Proposition 20
  

Isoceles Triangles

In today's blog, I show a very elementary property of isoceles triangles. This is one of the many proofs that I use to prove the existence of pi. Today's proof is taken straight from Euclid.

Definition 1: Isoceles Triangle

An isoceles triangle that has two sides of equal length.

Theorem 1: In an isoceles triangle, the base angles are congruent.




















Proof:

(1) Let triangle ABC be an isoceles triangle with AB ≅ AC

(2) Let D be a point that extends AB such that A,B,D are on the same line.

(3) Let E be a point that extends AC such that A,C,E are on the same line and AE ≅ AD.

(4) triangle DAC ≅ triangle EAB by Side-Angle-Side (see here) since:

(a) AB ≅ AC (step #1)

(b) ∠ DAC is a common angle.

(c) AE ≅ AD.

(5) We also know that triangle DBC ≅ triangle ECB by Side-Side-Side (see here if needed) since:

(a) BD ≅ CE since AD ≅ AG (step #3) and AB ≅ AC (step #1)

(b) BE ≅ CD (step #4) [By properties of congruent triangles, see here if needed]

(c) BC is a common side.

(6) So, now it follows that ∠ ABC ≅ ∠ ACB since:

(a) ∠ ABE ≅ ∠ ACD [By Properties of congruent triangles, see here if needed]

(b) ∠ EBC ≅ ∠ DCB [By Properties of congruent triangles and step #5]

(c) And finally, we know that:

∠ ABC = ∠ ABE - ∠ EBC

∠ ACB = ∠ ACD - ∠ DCB

QED

Corollary: If base angles are congruent, then sides are congruent










Proof:

(1) Let ABC be a triangle such that ∠ ABC ≅ ∠ ACB

(2) Assume that AB does not equal AC.

(3) Then one of them is greater. Let's assume AB. (Otherwise, we can make the same argument for side AC)

(4) Then there exists a point D such that BD is less than AB and BE ≅ AC

(5) From Theorem 1 above, we know that ∠ ABC ≅ ∠ DCB

(6) But this implies ∠ DCB ≅ ∠ ACB which is impossible.

(7) So we have a contradiction and we reject our assumption in #2.

QED

Theorem 2: The angle bisector of an isoceles triangle, is perpendicular to the base and divides up the base into two congruent segments.










Proof:

(1) Let AD be the angle bisector of ∠ BAC

(2) From this we, see that triangle BAD ≅ triangle CAD by S-A-S (see here) since:

(a) ∠BAD ≅ ∠ CAD since AD is the angle bisector.

(b) AB ≅ AC since ABC is an isoceles triangle.

(c) AD is a shared side between the two triangles.

(3) From congruent triangles (see here), we know that:

BD ≅ DC

∠ ADB ≅ ∠ ADC

(4) Now, since ∠ ADB, ∠ ADC are congruent and add up to 180 degrees (see here), we can conclude that they are both right angles.

QED

References

• David Joyce, Euclid's Elements, Book I, Proposition 5
  

Some properties of circles

In today's blog, I go over properties of a circle that I use later to prove the existence of pi. Today's proof is taken straight from Euclid.

Theorem 1: In a circle, if an angle that opens on the diameter, then it is a right angle.
















Proof:

(1) Let E be the center of the circle.

(2) BE ≅ BA ≅ CE since they are all radii.

(3) Since triangle AEB and triangle AEC are isoceles triangles (see here if needed), we can conclude (see here) that:

∠ ABE ≅ ∠ BAE

∠ ACE ≅ ∠ CAE

(4) And step #3 gives us that ∠ BAC = ∠ ABC + ∠ ACB.

(5) But we also know that ∠ FAC = ∠ ABC + ∠ ACB since:

(a) ∠ FAC = 180 degrees - BAC [Angles of a straight line add up to 180 degrees, see here if needed]

(b) ∠ ABC + ∠ ACB = 180 degrees - BAC [Angles in a triangle add up to 180 degrees, see here if needed]

(6) And since ∠ FAC ≅ ∠ BAC, both must be right angles [since 2*x = 180 degrees → x = 90 degrees]

QED

Postulate 1: Similar segments of circles on equal straight lines equal one another.

Euclid originally presented this postulate as a theorem using the principle of superposition (see here for details). I am presenting it as a postulate in order to avoid the superposition.

Lemma 1: In equal circles, angles stand on equal circumferences whether they stand at the centers or the circumferences.


























Proof:

(1) Let ABC and DEF be congruent circles with ∠ G ≅ ∠ H and ∠ A ≅ ∠ D.

(2) Since they are congruent, all radii are congruent so that:

BG ≅ CG ≅ EH ≅ FH

(3) So we have triangle BGC ≅ triangle EHF by side-angle-side (see here if needed)

(4) From step #3, we know that BC ≅ EF

(5) So that segment BAC ≅ segment EDF. [See Postulate I above]

(6) And this implies that segment BKC ≅ segment ELF.

QED

Lemma 2: In a circle, the angle at the center is double the angle at the circumference when the angles have the same circumference as base.






















Proof:

(1) Let ABC be a circle with center E.

(2) EA ≅ EB since both are radii.

(3) ∠ EAB ≅ ∠ EBA since the base angles of an isoceles triangle are congruent (see here for details if needed).

(4) ∠ BEF = ∠ EAB + ∠ EBA (since angles of a triangle add up to 180 degrees and since two angles of a straight line add up to 180 degrees) so ∠ BEF is double ∠ EAB.

(5) We can use the same line of reasoning to establish that ∠ FEC is double ∠ EAC.

(6) Putting step #4 and step #5 together gives us that ∠ BEC is double ∠ BAC

(7) We can use this same reasoning to prove that ∠ GEC is double ∠ EDC.

(8) We can also prove that ∠ GEB is double ∠ EDB.

(9) Therefore the remaining ∠ BEC is double ∠ BDC.

QED

Theorem 2: In equal circles, angles standing on equal circumferences are equal to one another, whether they stand at the center or at the circumference.




























Proof:

(1) So, we can assume that circumference BC ≅ circumference EC

(2) Assume ∠ BGC ≠ ∠ EHF

(3) Then one of them is greater; let's assume ∠ BGC is greater (if ∠ EHF is greater, we can make the same argument in terms of EHF)

(4) Construct ∠ BGK equal to ∠ EHF on the straight BG and at the point G on it (see here for details on the construction).

(5) Now equal angles stand on equal circumferences when they are at the centers, therefore circumference BK equals circumference EF (see Lemma 1 above)

(6) But EF equals BC so therefore BK equals BC which is a contradiction since BK is smaller than BC.

(7) So, we reject our assumption and conclude that ∠ BGC ≅ ∠ EHF

(8) The angle at A is half of the angle BGC. [See Lemma 2 above]

(9) The angle at D is half of the angle EHF [See Lemma 2 above]

(10) Therefore, the angle at A also equals the angle at D.

QED

Lemma 3: A line tangent to a point on a circle forms a right angle with a line drawn from that point to the center of the circle.

















Proof:

(1) Assume that ∠ FCD is not a right angle

(2) Let FG be a line that is perpendicular to DE

(3) So triangle FGC is a right angle with the hypotenuse at FC.

(4) So FC is greater than FG by the Pythaogrean Theorem [See the Corollary here for details]

(5) But FC = FB since radii are congruent.

(6) So FC is less than FG which contradicts step #4.

(7) So we have a contradiction and we reject our assumption.

QED

References:

• David Joyce, Euclid's Elements, Book III, Proposition 31
  

More on Equiangular Triangles

In today's blog, I review more proofs on equiangular triangles that are taken straight from Euclid's Elements. I use these properties in showing that the ratio of circumference to diameter for any circle is always the same (which is, of course, the definition of pi).

Lemma 1: If two triangles have one angle equal to one angle and the sides around the equal angle proportional, then the two triangles are equiangular.















Proof:

(1) Assume that ∠ BAC ≅ ∠ EDF and BA/AC = ED/DF

(2) There exists a point G such that ∠ FDG ≅ ∠ BAC and ∠ DFG ≅ ∠ ACB [See here for details on the construction.]

(3) ∠ B ≅ ∠ G [since the angles of a triangle add up to 180 degrees, see Lemma 4 here for details]

(4) triangle ABC is equiangular with triangle DGF [from step #2 and step #3]

(5) From the properties of equiangular triangles (see here), we know that:

BA/AC = GD/DF

(6) From our assumption in Step #1, we can conclude that:

GD/DF = ED/DF

(7) And from step #6, we can conclude that:

ED ≅ GD

(8) From Step #1 and Step #2, we can conclude that:

∠ EDF ≅ FDG

(9) We can now conclude that triangle DEF ≅ triangle DGF from Side-Angle-Side (see Postulate 1 here) since:

(a) DF ≅ DF

(b) ∠ EDF ≅ ∠ FDG [Step #8]

(c) ED ≅ GD [Step #7]

(10) ∠ ACB ≅ ∠ DFE since:

(a) ∠ ACB ≅ ∠ DFG [Step #2]

(b) ∠ DFG ≅ ∠ DFE [From Step #9, see here for properties of congruent triangles if needed]

(11) Finally, we can conclude that triangle ABC is equiangular with triangle DEF since:

(a) ∠ BAC ≅ ∠ EDF [By assumption in Step #1]

(b) ∠ ACB ≅ ∠ DFE [By Step #10]

(c) ∠ B ≅ ∠ E [Since angles of triangles add up to 180 degrees]

QED

Lemma 2: Those triangles which have one angle equal to one angle and which the sides about the equal angles are reciprocally proportional, are equal.













Proof:

(1) Let the sides of triangle ABC and triangle ADE be reciprocally proportional so that:

AE/AB = CA/AD

(2) Since triangle ABC and triangle ABD share the same height, we can conclude that (see here):

AC/AD = area triangle ABC/area triangle ABD

(3) Likewise, since triangle ADE and triangle ABD share the same height, we can conclude that:

AE/AB = area triangle ADE/triangle ABD

(4) Therefore, triangle ABC/triangle ABD = triangle ADE/triangle ABD.

(5) But then we have:

area triangle ABC * area triangle ABD = area triangle ADE * area triangle ABD

(6) And if we divide both sides by the area of triangle ABD, we get:

area triangle ABC = area triangle ADE.

QED

Lemma 3: Similar triangles are one to another in the square ratio of corresponding sides.
















Proof:

(1) Let triangle ABC and triangle DEF be equiangular triangles such that:

∠ A ≅ ∠ D
∠ B ≅ ∠ E
∠ C ≅ ∠ F

(2) There exists a point G such that BG = [(EF)*(EF)]/BC

(3) From (#2) we have:

1/BG = BC/[(EF)*(EF)]

which implies that:

EF/BG = BC/EF

(4) From a Property of Equiangular Triangles (see here if needed), we know that:

AB/BC = DE/EF

so that:

AB/DE = BC/EF

(5) We can also conclude that triangle ABG has equal area to triangle DEF since:

(a) ∠ B ≅ ∠ E (step #1)

(b) AB/DE = EF/BG (from combining step #3 with step #4)

(c) Lemma 2 above

(6) We can also conclude that BC/BG = (CB)²/(EF)² since:

BC*BG = EF*EF (step #2)

And if we multiply BC to both sides, we get:

(BC)²*BG = (EF)²*BC

If we divide BG from both sides and (EF)² from both sides, we get:

(BC)²/(EF)² = BC/BG

(7) Since triangle ABC and triangle ABG have the same height, we can conclude (see here):

CB/BG = area triangle ABC/area triangle ABG

(8) Applying setp #6, gives us:

area triangle ABC/area triangle ABG = (BC)²/(EF)²

(9) Finally, from step #5, we have:

area triangle ABC/area triangle DEF = (CB)²/(EF)²

QED

References

• David Joyce, Euclid's Elements
  

derivate-of-sine-and-cosine

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Euclid and pi

Pi, also know as Archimede's constant, is not mentioned in Euclid's Elements. The closest that Euclid comes is Proposition II in Book XII which states that two circles are to each other as the squares of their diameters.

Postulate 1: Law of Trichotomy

For any two values x,y, there are only three possible states:
(a) x = y
(b) x is less than y
(c) x is greater than y

This is one of the postulates of real numbers. See here for details on constructing real numbers.

Lemma 1: Similar polygons inscribed in circles are to one another as the squares on their diameters





























Proof:

(1) By assumption, polygon ABCDE is similar to polygon FGHKL with BM and GN being the diameters of circles.

(2) From the property of similar polygon (see definition above), we have:

∠ BAE ≅ ∠ GFL

BA/AE = GF/GL

(3) triangle ABE is equiangular to triangle FGL [See Lemma 1 here for details]

(4) Since they are equiangular, we know that:

∠ AEB ≅ ∠ FLG

(5) Since both angles open on the same length of the circumference (see here):

∠ AEB ≅ ∠ AMB
∠ FLG ≅ ∠ FNG

(6) From (4) and (5), we can conclude that:

∠ AMB ≅ ∠ FNG

(7) Since BM and GN are both diameters of the circle (see here),
we can conclude that both ∠ BAM and ∠ GFN are right angles.

(8) Since the angles of a triangle add up to 180 degrees (see Lemma 4 here), we can conclude from step #6 and step #7 that triangle ABM is equiangular to triangle FGN.

(9) From the properties of equiangular triangles (see Lemma 3 here), we know that:
BM/GN = BA/GF

(10) From (9), we can conclude that:
(BM/GN)² = (BA/GF)² = BM²/GN² = BA²/GF²

(11) From similar polygons (See Theorem here), we know that:
(Area of ABCDE)/(Area of FGHKL) = BA²/GF²

(12) Putting this all together, gives us:
BM²/GN² = (Area of ABCDE)/(Area of FGHKL)

QED

Lemma 2: if A/B = C/D with A greater than C, then D is less than B.

Proof:

(1) Let A/B = C/D with A greater than C.

(2) So that AD = BC

(3) Now, D ≠ B since if D = B, then AD is greater than BC which contradicts step #2.

(4) Now, D cannot be greater than B since then AD is greater than BC which contradicts step #2.

(5) So, by the Law of Trichotomy (see Postulate above), we can conclude that D is less than B.

QED


Theorem: Two circles are to each other as the squares of their diameters.













































Proof:

(1) Let C₁ be the circle formed with diameter BD and area A₁.

(2) Let C₂ be the circle formed with diameter FH and area A₂.

(3) Assume that A₁/A₂ ≠ (BD)²/(FH)²

(4) There exists an area S such that: (BD)²/(FH)² = A₁/S

(5) Assume that S is less than A₂

(6) Then, there exists a polygon EKFLGMHN such that the area of this polygon is greater than the area of S. [From the Method of Exhaustion, see Lemma 2.]

(7) We can inscribe a similar polygon into circle C₁ [See here for details on this construction]

(8) From Lemma 1 above, we can conclude:

BD²/FH² = (Area polygon AOBPCQDR)/(Area polygon EKFLGMHN)

(9) But then, from step #4:

BD²/FH²= A₁/S

(10) So we can conclude that:

(Area polygon AOBPCQDR)/(Area polygon EKFLGMNH) = A₁/S

(11) Since A₁ is greater than Area polygon AOBPCQDR, we can conclude from step #15 that S is greater than Area polygon EKFLGMN from Lemma 3 above.

(12) But this is impossible since in step #6 we showed that S is less than this same regular polygon so we have a contradiction and we reject our assumption in step #5.

(13) Now, let's assume that S is greater than A₂

(14) So this means that (FH)²/(BD)² = S/A₁

(15) Let T be the area such that S/A₁ = A₂/T

(16) We can see that T is less than A₁ from Lemma 2 above.

(17) There exists a regular polygon that is greater in area than T but smaller than the area of the circle by the Method of Exhaustion (see Lemma 2)

(18) We can inscribe a similar polygon in A₂.

(19) From Lemma 1 above, we can conclude:

FH²/BD² = (Area polygon EKFLGMHN)/(Area polygon AOBPCQDR)

(20) Likewise from step #14, we have:
(FH)²/(BD)² = S/A₁

(21) And from step #15, this means that:

(Area polygon EKFLGMNH)/(Area polygon AOBPCQDR) = A₂/T

(22) Now A₂ is greater in area than polygon EKFLGMNH so that T must be greater than the area of polygon AOBPCQDR

(23) But this is impossible from step #17 so we have a contradiction and we reject step #13.

(29) We now apply the Law of Trichotomy (see Postulate above) and we are done.

QED

References

• Phil Schultz, Eudoxus's Proof that the area of a circle is a constant times diameter squared
• David Joyce, Euclid's Elements
  

Derivative of sine and cosine

In today's blog, I bring together some of the results that I presented earlier to determine the derivative for sine and cosine.

Lemma 1: sin(A+B) - sin(A - B) = 2*cos(A)sin(B)

Proof:

(1) sin(A+B) = cos(B)*sin(A) + cos(A)*sin(B) [See here for proof]

(2) sin(A-B) = sin(A+(-B)) = cos(-B)*sin(A) + cos(A)*sin(-B) =

(3) since cos(-B) = cos(B) [See here] and sin(-B) = -sin(B) [See here], we get:

sin(A-B) = cos(B)*sin(A) - cos(A)*sin(B)

(4) sin(A+B) - sin(A-B) =

= cos(B)*sin(A) + cos(A)*sin(B) - cos(B)*sin(A) + cos(A)*sin(B) =

= 2*cos(A)*sin(B)

QED

Lemma 2: sin P - sin Q = 2*cos [(P+Q)/2 ] * sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A+B = (P+Q)/2 + (P-Q)/2 = (P+P+Q-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P-P+Q+Q)/2 = Q

(5) sin(P) - sin(Q) = sin(A+B) - sin(A-B) = 2*cos(A)*sin(B) [See Lemma 1 above]

(6) Putting it all together gives us:

sin(P) - sin(Q) = 2*cos(A)*sin(B) = 2*cos[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 1: d/dx(sin x) = cos x

Proof:

(1) Let y = sin(x)

(2) dy/dx = lim (Δx → 0) [sin(x + Δx) - sin(x)]/Δx

(3) From Lemma 2 above we know that:
sin(x+Δx) - sin(x) = 2*cos[(x+Δx+x)/2]*sin[(x+Δx-x)/2] =
= 2*cos(x + Δx/2)*sin(Δx/2)

(4) So, substituting (3) into (2) gives us:

dy/dx = lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ]

(5) Using the Product Law (see here), we know that:

lim(Δx → 0) [ 2*cos(x+Δx/2)*sin(Δx/2) ] =
lim(Δx → 0)[cos(x+Δx/2)] * lim(Δx → 0)[sin(Δx/2)/(Δx/2)]

(6) Now, if we set θ = Δx/2, we know that:
lim (θ → 0) [ sin(θ)/θ ] = 1 (See here for proof)

(7) We also know that
lim(Δx → 0)[cos(x + Δx/2)] = cos x

(8) This then gives us:
dy/dx = cos x * 1 = cos x.

QED

Lemma 3: cos(A+B) - cos(A-B) = -2*sin(A)sin(B)

Proof:

(1) cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) [See here for proof]

(2) cos(A-B) = cos(A+(-B)) = cos(A)*cos(-B)-sin(A)*sin(-B)

(3) Since cos(-B) = cos(B) and sin(-B) = -sin(B) [See here], we have:

cos(A-B) = cos(A)*cos(B)+sin(A)*sin(B)

(4) cos(A+B) - cos(A-B) =

=cos(A)*cos(B) - sin(A)*sin(B) - cos(A)*cos(B) - sin(A)*sin(B) =

= -2*sin(A)*sin(B)

QED

Lemma 4: cos P - cos Q = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

Proof:

(1) Let A = (P+Q)/2

(2) Let B = (P-Q)/2

(3) A + B = (P+Q)/2 + (P-Q)/2 = (P+Q+P-Q)/2 = P

(4) A - B = (P+Q)/2 - (P-Q)/2 = (P + Q - P + Q)/2 = Q

(5) cos(P) - cos(Q) = cos(A+B) - cos(A-B) = -2*sin(A)*sin(B) [From Lemma 3 above]

(6) So,

cos(P) - cos(Q) = -2*sin(A)*sin(B) = -2*sin[(P+Q)/2]*sin[(P-Q)/2]

QED

Theorem 2: d/dx(cos x) = -sin x

Proof:

(1) Let y = cos(x)

(2) dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx

(3) Now, from Lemma 4 above:

cos(x +Δx) - cos(x) = -2*sin[(x + Δx + x)/2]*sin[(x+Δx-x)/2] =
= -2*sin(x + Δx/2)*sin(Δx/2)

(4) Once again, applying the Product Rule for limits gives us:

dy/dx = lim(Δx → 0) [ cos(x + Δ x) - cos(x) ]/Δx =
lim(Δx → 0)[ -sin(x + Δx/2) ] * lim(Δx → 0) [ sin (Δx/2)/(Δx/2)]

(5) Again, setting θ = Δx/2 gives us:
lim(Δx → 0)[ sin(θ)/θ ] = 1 [See here for proof]

(6) We can also see that:
lim(Δx → 0)[ -sin(x + Δ x/2) ] = -sin(x)

(7) Putting this all together gives us:
dy/dx = -sin(x)*1 = -sin(x)

QED

limit (θ → 0) sin θ/θ = 1

Today's proof is part of the review of basic properties that I use to determine the derivatives of sine x and cosine x. This is part of the larger story where I show how the Taylor Series can be used to define sin and cosine independently of Euclid.

To be clear, I am using Euclidean Geometry to determine the derivative for sin x and cosine x, then I am using the Taylor Series to show that these results are equivalent to an infinite series that makes no such assumption. These ideas form the foundation of Euler's Identity which is one of the most amazing results in all of mathematics.

Lemma 1: if (a/b) is greater than (c/d), then d/c is greater than b/a

Proof:

(1) Let n be a positive value such that 10ⁿ that is greater than (a/b).

(2) 10ⁿ / (c/d) is greater than 10ⁿ/(a/b) since any number can be divided more times by a smaller amount.

(3) But now, if we divide both sides of the equation by 10ⁿ, we get:

1/(c/d) is greater than 1/(a/b)

which means that:

d/c is greater than b/a

QED

Lemma 2: lim(θ → 0) sin θ/θ = 1



Proof:

(1) Let r be the length of the radius of circle O.

(2) The area for triangle OAB = (1/2)r*(r*sin θ) [See here for details if needed]

= (1/2)r²*sin(θ)

(3) The area of the sector OAB = (1/2)r²θ [Proof to be added later]

(4) The area of the triangle OAT = (1/2)r²tan(θ)

[Since the area of a triangle is (1/2)base*height, see here if needed, with base = r and height = r*tan(θ), see here if needed]

Now, since tan(θ) = sin(θ)/cos(θ) [see here if needed], then we have:

The area of triangle OAT = (1/2)r²(sin θ)/(cos θ)

(5) So, we can see that:

area of triangle OAB is less than area of the sector OAB which is less than area of the triangle OAT.

So that:

(1/2)r²(sin θ) is less than (1/2)r²θ which is less than (1/2)r²(sin θ)/(cos θ)

(6) Dividing all sides by (1/2)r² gives us:

sin θ is less than θ which is less than sin(θ)/cos(θ)

(7) Now, if we divide (5) by sin θ (assuming sin θ ≠ 0), then we get:

1 is less than θ/(sin θ) which is less than 1/cos(θ)

Taking the reciprocal for each value gives us:

1 is greater than (sin θ/θ) which is greater than cos(θ).

(8) Now, we know that the limit (θ → 0) 1 = 1, by the Constant Law (see here).

(9) We know that lim(θ → 0) cos(θ) = 1 since:

(a) cos θ is a continuous function (see here if more details are needed)

(b) so this means lim (θ → 0) cos θ = cos 0 = 1 (see here for definition of continuous functions, see here for review of why cos 0 = 1)

(10) But now, we can apply the Squeeze Rule (see here), to get:

lim (θ → 0) (sin θ/θ) = 1.

QED

sin(a+b) and cos(a+b)

In today's blog, I plan to go over a basic property of sine and cosine that I will use later to find the derivatives of sine and cosine.

This is part of the larger story of using Taylor Series to come up with an equation for sine and cosine that does not depend on Euclid.

Lemma 1: Area of a triangle = (1/2)(ab)(sin C)














Proof:

Case I: C is less than 90 degrees.

(1) Let AD be perpendicular to BC

(2) We know that the area of triangle ABC is (1/2)ah (where a = the length of BC) [See here for details if needed]

(3) Now, we also know that:

sin C = h/b [Definition of sine, see here]

So that:

h = b sin C

(4) This gives us:

Area of triangle ABC = (1/2)ab*sin C [By substituting (#3) for (#2)]

Case II: C = 90 degrees

(1) Sine 90 degrees = 1 (see here for details if needed)

(2) In this case, b = h.

(3) So, the area is (1/2)ah = (1/2)ab = (1/2)ab sin 90 degrees = (1/2)ab sin C

Case III: C is greater than 90 degrees














(1) Let AD be perpendicular to BC so that B,C,D are colinear.

(2) We know that the area of triangle ABC is (1/2)ah (where a = the length of BC) [See here for details if needed]

(3) Now, we also know that:

sin C = h/b [Definition of sine, see here; since sin C = sin (180 - C).

So that:

h = b sin C

(4) This gives us:

Area of triangle ABC = (1/2)ab*sin C [By substituting (#3) for (#2)]

QED

Theorem 1: sin(a+b) = cosBsinA + cosAsinB



Proof:

(1) Let QN be a line that is perpendicular with PR.

(2) The area of triangle PQR = area of triangle PQN + area of triangle RQN

(3) By Lemma 1 above, we have:

Area of triangle PQR = (1/2)rp*sin(A+B)

Area of triangle PQN = (1/2)rh*sin(A)

Area of triangle RQN = (1/2)ph*sin(B)

(4) Combing step #2 with step #3 gives us:

(1/2)rp*sin(A+B) = (1/2)rh*sin(A) + (1/2)ph*sin(B)

(5) Multiplying 2 to both sides gives us:

rp*sin(A+B) = rh*sin(A) + ph*sin(B)

(6) Dividing both sides by rp gives us:

sin(A+B) = (h/p)*sin(A) + (h/r)*sin(B)

(7) Since QN is perpendicular to PR, we also know that (see here for definition of cosine):

cos B = h/p

cos A = h/r

(8) So that we have:

sin(A+B) = cos(B)*sin(A) + cos(A)sin(B)

QED

Lemma 2: Law of Cosines

c² = a² + b² - 2ab*cos c

Proof:

Case I: Obtuse Angle (a triangle with one angle greater than 90 degrees)














(1) Let C be the obtuse angle (by C, I mean ∠ ACB)

(2) Let d = a + e

(3) Using Pythagorean Theorem (see here):

c² = d² + h²
b² = e² + h²

(4) So, we have:

c² = (a+e)² + h² = a² + 2ae + e² + h² =
= a² + 2ae + b²

(5) Now, cos C = -cos(180-C) = -e/b [See here for details if needed]

So, we also have that:

e = -b*(cos C)

(6) Combining step #5 with step #6 gives us:

c² = a² + b² - 2ab*(cos C).

Case II: Right triangle

In this case, the Pythagorean Theorem applies and we have:

c² = a² + b²

Now cos 90 degrees = 0 (see here if needed), so we also have:

c² = a² + b² - 2ab*(cos C)

Case III: Acute triangle





(1) Let c be the acute angle.

(2) We will need to consider two cases. One where B is an acute angle and one where B is an obtuse angle.

(3) Let h be the height of the triangle.

(4) Let e represent the segment CD.

(5) Using Pythagorean Theorem, we have:

c² = d² + h²

b² = e² + h²

(6) Let a represent the segment BC.

(7) We see that in both cases,

d² = (e-a)² since:

(a) For the first triangle, a = d + e which implies -d = e - a.

(b) For the second triangle, e = d + a which implies d = e - a.

(8) And from this, we have:

d² = e² - 2ae + a²

which gives us:

d² - e² = a² - 2ae

(9) From step #5, we have:

h² = b² - e²

So we also have:

c² = d² + h² = d² + b² - e² =
= a² - 2ae + b²

(10) Now, cos C = e/b [See here for definition of cosine]

So we have: e = b cos C

And this then gives us:

c² = a² + b² - 2ab(cos C)

QED


Theorem 2: cos(A+B) = cosA*cosB - sinA*sinB

NOTE: I will use the same diagram as I used for Theorem 1

Proof:

(1) (x+y)² = r² + p² - 2rpcos(A+B) [See Law of Cosines above]

(2) If we substract (x+y)² from both sides and add 2rp*cos(A+B) to both sides, we get:

2rp*cos(A+B) = r² + p² - (x+y)² =

= r² + p² -x² - 2xy - y² =

= (r² - x²) + (p² - y²) - 2xy

(3) From the Pythagorean Theorem (see here), we get:

r² = h² + x²

and

p² = h² + y²

so that:

h² = r² - x²

and

h² = p² - y²

(4) Combining the results of step #3 with step #2 gives us:

2rp*cos(A+B) = r² + p² - (x+y)² =(h² + x²) + (h² + y²) + 2xy - x² - y² = 2h² - 2xy

(5) Dividing both sides by 2*rp gives us;

cos(A+B) = h²/(rp) - (xy)/(rp)

(6) Now from the definitions of sine and cosine (see here), we have:

cos A = h/r

cos B = h/p

sin A = x/r

sin B = y/p

(7) Combinining step #6 with step #5 gives us:

cos (A+B) = cosA*cosB - sinA*sinB

QED

References

• David Joyce, Area of a Triangle
• David Joyce, Law of Cosines
• Davis Associates, Pythagorean Theorem