Lecture 25: The Sine and Cosine Functions. tan(x) 1+y

Transcription

1 Lecture 5: The Sine Cosine Functions 5. Denitions We begin b dening functions s : c : ; i! R ; i! R b Note that 8 >< q tan(x) ; if x s(x) + tan (x) ; >: ; if x 8 >< q ; if x c(x) + tan (x) ; >: 0; if x. s(x) x!!+ +!+ q + c(x) x!!+ +!+ which shows that both s c are continuous functions. Next we extend the denitions of s c to functions q + 0; S : ; 3 i! R C : ; 3 i! R b dening 8 >< s(x); if x ; i S(x) >: s(x ); if x ; 3 i 5-

2 Lecture 5: The Sine Cosine Functions 5-8 >< c(x); if x ; i C(x) >: c(x ); if x ; 3 i. Note that S(x) s(x) x! + x! + C(x) x! + x! + c(x)!! + +!! q+ q + 0; which shows that both S C are continuous at.thus both S C are continuous. Finall for an x R let g(x) sufn : n Z; +n < xg: Denition With the notation as above for an x R we call the sine cosine of x resectivel. S(x g(x)) cos(x) C(x g(x)) Proosition The sine cosine functions are continuous on R. From the denitions it is sucient toverif continuit at 3.Now x! 3 x! 3 S(x) 3 S s x! 3 + x! 3 S(x ) + x! + s(x)! +! q + ;

3 Lecture 5: The Sine Cosine Functions 5-3 so sine is continuous at 3. Similarl x! 3 cos(x) x! 3 C(x) 3 C c 0 x! 3 + cos(x) x! 3 C(x ) + c(x) x! +! +! 0; q + so cosine is continuous at Proerties of sine cosine Proosition The sine cosine functions are eriodic with eriod. Proosition The result follows immediatel from the denitions. For an x R cos(x) cos(x). The result follows immediatel from the denitions. Proosition For an x R sin (x) + cos (x). The result follows immediatel from the denition of s c. Proosition The range of both the sine cosine functions is [; ]. The result follows immediatel from the denitions along with the facts that + jj + for an R. Proosition For an x in the domain of the tangent function tan(x) cos(x) : The result follows immediatel from the denitions.

4 Lecture 5: The Sine Cosine Functions 5-4 Proosition For an x in the domain of the tangent function sin (x) tan (x) + tan (x) cos (x) + tan (x) : The result follows immediatel from the denitions. Proosition For an x; R cos(x + ) cos(x) cos() sin(): First suose x x + are in the domain of the tangent function. Then cos (x + ) + tan (x + ) tan(x) + tan() + tan(x) tan() ( tan(x) tan()) ( tan(x) tan()) + (tan(x) + tan()) ( tan(x) tan()) ( + tan (x))( + tan ()) + tan (x) + tan ()! tan(x) tan() + tan (x) + tan () (cos(x) cos() sin()) : Hence cos(x + ) (cos(x) cos() sin()): Consider a xed value of x. Note that the ositive sign must be chosen when 0. Moreover increasing b changes the sign on both sides so the ositive sign must be chosen when is an multile of. Since sine cosine are continuous functionsthe

5 Lecture 5: The Sine Cosine Functions 5-5 choice of sign could change onl at oints at which both sides are 0 but these oints are searated b a distance of so we must alwas choose the ositive sign. Hence we have cos(x + ) cos(x) cos() sin() for all x; R for which x x + are in the domain of the tangent function. The identit for the other values of x now follows from the continuit of the sine cosine functions. Proosition For an x; R sin(x + ) cos() + sin() cos(x): Exercise 5.. Prove the revious roosition. Exercise 5.. Show that for an x R sin x cos(x) cos x : Exercise 5..3 Show that for an x R Exercise 5..4 Show that for an x R Exercise 5..5 Show that cos(x) cos(x) cos (x) sin (x): sin (x) cos(x) cos (x) + cos(x) : sin cos ; 4 4 sin cos 6 3 ; sin cos :

6 Lecture 5: The Sine Cosine Functions The calculus of the trigonometric functions arctan(x) Proosition. Using l'h^oital's rule arctan(x) x!0 +x : tan(x) Proosition. Letting x arctan(u) we have Proosition We have tan(x). cos(x) Proosition We have Proosition We have 0. cos(x) u u!0 arctan(u) : tan(x) cos(x) : x!0 cos(x) + cos(x) x + cos(x) cos (x) ( + cos(x)) x!0 x ()(0) 0: If f(x) then f 0 (x) cos(x). + cos(x) f 0 sin(x + h) (x) h!0 h cos(h) + sin(h) cos(x) h!0 h cos(h) sin(h) + cos(x) h!0 h h!0 h cos(x):

7 Proosition Exercise 5.3. Prove the revious roosition. Lecture 5: The Sine Cosine Functions 5-7 If f(x) cos(x) then f 0 (x). Denition For aroriate x R cot(x) cos x ; sec(x) cos(x) ; csc(x) are called the cotangent secant cosecant of x resectivel. Exercise 5.3. If f(x) tan(x) g(x) cot(x) h(x) sec(x) k(x) csc(x) show that f 0 (x) sec (x); g 0 (x) csc (x); h 0 (x) sec(x) tan(x); k 0 (x) csc(x) cot(x): Z Proosition x dx. Let x sin(u). Then as u varies from to x varies from to. And for these values wehave q x sin (u) cos (u) jcos(u)j cos(u): Hence Z x dx Z Z Z cos (u)du + cos(u) du + Z du cos(u)du + (sin() sin()) 4 :

8 Lecture 5: The Sine Cosine Functions 5-8 Exercise Find the Talor olnomial P 9 of order 9 for f(x) at 0. Note that this is equal to the Talor olnomial of order 0 for f at 0. Is P 9 an overestimate or an underestimate for sin? Find an uer bound for the error in this aroximation.