Continuity and Differentiability of the Trigonometric Functions

Transcription

1 [Te basis for te following work will be te definition of te trigonometric functions as ratios of te sides of a triangle inscribed in a circle; in particular, te sine of an angle will be defined to be te ratio of te triangle s opposite side to its ypotenuse - wic ere simplifies to being te triangle s opposite side, as noted in te work. I make no reference to te series definitions of te trigonometric functions. Angles are given in radian measure. Te eplicit statement of te teorem proffered as data is given by Euclid VI 33.] Continuity and Differentiability of te Trigonometric Functions It may be taken as data tat te ratio of te area of a circle to tat (A s ) of a sector terein is equal to te ratio of te circumference of te circle to te lengt (θ s r) of te sector s arc; and tus A s πr = θ sr πr A s = θ sr (1) Below is a geometric figure, illustrating a comparison of areas (wic will prove useful.) 1

2 We consider a unit circle, restricting our attention (for now) to angles suc tat 0 < π. As te figure illustrates, for a sector described by an angle in tis range, te area of an inscribed triangle (wose eigt is, and wose base as unit lengt) is less tan or equal to te area of te sector. But te area of te triangle must be given by, wereas (1) indicates tat te area of te sector is given by ; and from tis, it follows tat, 0 < π From te trigonometric identity (sin( ) = -), it follows similarly tat for values of suc tat π < 0, is larger tan or equal to (using te identity, = ( ()) = (sin( )); but for in tis range, we know already tat sin( ) ; so sin( ) = ). In sort, we find 0, 0 < π () 0, π < 0 (3) But from tese te Squeeze Teorem tells us, since = 0, tat 0 = 0. (In detail, we must take eac of te one-sided its as approaces zero; since tese eist and are zero, te two-sided it eists and is zero.) From tis we may te prove te fact, wic we will sortly use, tat 0 cos() = 1. Proof : Fi ɛ > 0, and set δ 1 = ɛ. Our foregoing work indicates tat δ : sin(t) < ɛ if 0 < t < δ In particular, if we consider satisfying < δ, ten - since < generally - we know tat for any on te interval, it is true tat Terefore, for te δ in te it statement sin( ) < ɛ if 0 < < δ

3 cos() 1 < ɛ if 0 < < δ, coose δ = min(δ 1, δ ). Ten < δ < ɛ, and tus < ɛ ; and our selection of δ also implies tat sin ( ) < < ɛ, so tat sin ( ) < ɛ. But te trigonometric identity (1 cos() = sin ( )) indicates tat we ave sin ( ) = cos() 1 < ɛ wic sows tat our δ is satisfactory. Wit tese its in and, we may additionally determine two more its related to te trigonometric functions wic will be key to later work. Tese are 0 1 cos() 0 = 1 (4) = 0 (5) Te proof of (4) as a geometric derivation wic epands upon te derivation of 0. 3

4 Again inscribing a triangle in te unit circle, we etend tis triangle so tat te terminal point of its ypotenuse lies above tat of te radial ais of te angle s measure, α. Tus te lengt of te opposite side in tis triangle is tan(). Note tat te area of te sector falls in between and tan() (tese being te areas of te triangles inscribed in te circle, and engulfing it, respectively). Terefore, for satisfying 0 < < π, < < tan() Multiplying by, we find 1 < < 1 cos() and, on taking reciprocals, we arrive at 1 > > cos() To obtain a symmetric inequality for satisfying π function; since we already ave < < 0, note tat tan() is an odd sin( ) < < tan( ) from our prior work, we take te negative of eac constituent inequality, and find > > tan() Since < 0 in tis contet, te step of multiplying by once more: reverses te inequalities 1 < < 1 cos() 4

5 and terefore taking reciprocals leaves us wit te identical result. Tus, 1 > > cos() if 0 < < π Now - to make eplicit ow te Squeeze Teorem may be applied ere to derive (4) - we may fi ɛ > 0. For tis ɛ, since cos() = 1, we know tat tere eist δ 1, δ > 0 suc tat 0 < ɛ if 0 < < δ 1 and cos() 1 < ɛ if 0 < < δ. Coosing δ = min(δ 1, δ, π ), we find tat if 0 < < δ, ten 1 > > cos(), and so 1 + ɛ > 1 > > cos() > 1 ɛ. Terefore 1 < ɛ if 0 < < δ. Tis proves (4). (5) follows easily: 1 cos() 0 (1 cos())(1 + cos()) 0 ()(1 + cos()) = (1)( cos() ) = 0 0 sin () ()(1 + cos()) We are now ready to prove te following teorems: (1) Teorem: Te si trigonometric functions are continuous on teir natural domains. () Teorem: Te si trigonometric functions are differentiable on teir natural domains. Proof of (1): We sow first tat is continuous everywere. We say tat is continuous at 0 if = sin( 0 ) 0 If we put = 0, ten tis is equivalent to sin( 0 + ) = sin( 0 ) 5

6 But, by te trigonometric identity for sin(α + β), and by our previous work, 0 + ) [sin( 0 ) cos() + sin() cos( 0 )] sin( 0 ) cos() + sin() cos( 0 ) = (sin( 0 ))(1) + (0)(cos( 0 )) = sin( 0 ) Tus is continuous for any arbitrary value in its domain, since te identity and it statements wic we used old true everywere in its domain. Since te natural domain of includes all real values (our definition of indicates tat it is a periodic function), is terefore continuous everywere. Second, we sow tat cos() is continuous everywere. As wit, we may define te condition for continuity of cos() at 0 by our variable ; i.e., cos( 0 + ) = cos( 0 ) implies continuity at tis point. Using te trigonometric identity for cos(α + β), and our previous work, 0 + ) [cos( 0 ) cos() sin() sin( 0 )] cos( 0 ) cos() sin() sin( 0 ) = (cos( 0 ))(1) (0)(sin( 0 )) = cos( 0 ) Given te identically periodic nature (in relation to ) of cos(), we conclude similarly tat cos() is continuous everywere. Finally, we note tat, since te oter trigonometric functions can be defined as ratios or reciprocals of and cos(), te continuity of and cos() guarantees tat tese ratios are also continuous were defined; and te equivalence of tese definitions wit our original ones (based upon triangles inscribed in circles) implies tat te natural domain of tese functions is coincident wit were tese ratios are defined. Terefore, we conclude tat tese functions - and tus, all trigonometric functions - are continuous on teir natural domains. 6

7 Proof of (): We sow only tat and cos() are differentiable everywere; it ten follows unambiguously from te Quotient Rule for derivatives tat te oter trigonometric functions are differentiable on teir natural domains. (Teir formulaic derivatives may be determined by use of te Quotient Rule as well.) For, te derivative eists and is given by sin( + ) cos() + sin() cos() cos() sin() cos() + 1 cos() = = cos() + cos() sin() and for cos(), cos( + ) cos() cos() cos() sin() cos() cos() cos() cos() sin() 1 cos() = cos() = sin() and terefore bot and cos() are differentiable. 7