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Section 3.4
      Derivatives of Trigonometric Functions

                           Math 1a


                      February 25, 2008


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Two important trigonometric limits




   Theorem
   The following two limits hold:
            sin θ
        lim        =1
       θ→0 θ
            cos θ − 1
        lim           =0
       θ→0       θ
Proof of the Sine Limit
                                  Proof.
                                  Notice

                                           sin θ ≤ θ ≤ tan θ

                                  Divide by sin θ:
                                                  θ       1
                                        1≤            ≤
                                                sin θ   cos θ
                                  Take reciprocals:
                  sin θ θ tan θ
          θ                                     sin θ
                                           1≥         ≥ cos θ
          cos θ         1                         θ
                                  As θ → 0, the left and right
                                  sides tend to 1. So, then,
                                  must the middle
                                  expression.
Now
      1 − cos θ   1 − cos θ 1 + cos θ        1 − cos2 θ
                =           ·           =
         θ            θ       1 + cos θ     θ(1 + cos θ)
                        2
                     sin θ        sin θ     cos θ
                =               =       ·
                  θ(1 + cos θ)      θ     1 + cos θ

So
          1 − cos θ          sin θ             cos θ
      lim           = lim            ·   lim
      θ→0    θ          θ→0 θ            θ→0 1 + cos θ

                    = 1 · 0 = 0.
Derivatives of Sine and Cosine
   Theorem
                         d
                            sin x = cos x.
                         dx

   Proof.
   From the definition:
Derivatives of Sine and Cosine
   Theorem
                            d
                               sin x = cos x.
                            dx

   Proof.
   From the definition:
            d              sin(x + h) − sin x
               sin x = lim
            dx         h→0         h
Derivatives of Sine and Cosine
   Theorem
                             d
                                sin x = cos x.
                             dx

   Proof.
   From the definition:
            d              sin(x + h) − sin x
               sin x = lim
            dx         h→0          h
                           (sin x cos h + cos x sin h) − sin x
                     = lim
                       h→0                 h
Derivatives of Sine and Cosine
   Theorem
                              d
                                 sin x = cos x.
                              dx

   Proof.
   From the definition:
            d               sin(x + h) − sin x
               sin x = lim
            dx         h→0           h
                            (sin x cos h + cos x sin h) − sin x
                     = lim
                       h→0                  h
                                   cos h − 1                 sin h
                     = sin x · lim           + cos x · lim
                               h→0      h               h→0 h
Derivatives of Sine and Cosine
   Theorem
                              d
                                 sin x = cos x.
                              dx

   Proof.
   From the definition:
            d               sin(x + h) − sin x
               sin x = lim
            dx         h→0           h
                            (sin x cos h + cos x sin h) − sin x
                     = lim
                       h→0                   h
                                   cos h − 1                 sin h
                     = sin x · lim            + cos x · lim
                               h→0      h               h→0 h
                     = sin x · 0 + cos x · 1 = cos x
Illustration of Sine and Cosine




                       y


                                          x
       π        π      0          π   π
              −
                2                 2       sin x
Illustration of Sine and Cosine




                       y


                                           x
       π        π      0          π   π
              −                           cos x
                2                 2       sin x
Derivatives of Sine and Cosine




   Theorem
       d
          sin x = cos x.
       dx
       d
          cos x = − sin x.
       dx
Derivatives of tangent and secant



   Fill in the table:
                          y       y
                        sin x   cos x
                        cos x   − sin x
                        tan x
                        cot x
                        sec x
                        csc x
Derivatives of tangent and secant



   Fill in the table:
                          y       y
                        sin x   cos x
                        cos x   − sin x
                        tan x   sec2 x
                        cot x
                        sec x
                        csc x
Derivatives of tangent and secant



   Fill in the table:
                          y        y
                        sin x    cos x
                        cos x   − sin x
                        tan x    sec2 x
                        cot x   − csc2 x
                        sec x
                        csc x
Derivatives of tangent and secant



   Fill in the table:
                          y         y
                        sin x     cos x
                        cos x     − sin x
                        tan x     sec2 x
                        cot x    − csc2 x
                        sec x   sec x tan x
                        csc x
Derivatives of tangent and secant



   Fill in the table:
                          y          y
                        sin x      cos x
                        cos x      − sin x
                        tan x      sec2 x
                        cot x     − csc2 x
                        sec x    sec x tan x
                        csc x   − csc x cot x
Example
Find
                sin 2θ
          lim
          θ→0      θ
Example
Find
                                     sin 2θ
                               lim
                               θ→0      θ

Solution (i)
Use a trig identity:
    sin 2θ       2 sin θ cos θ           sin θ
lim        = lim               = 2 · lim       · lim cos θ = 2 · 1 · 1 = 1
θ→0    θ     θ→0       θ            θ→0 θ       θ→0
Example
Find
                                     sin 2θ
                               lim
                               θ→0      θ

Solution (i)
Use a trig identity:
    sin 2θ       2 sin θ cos θ           sin θ
lim        = lim               = 2 · lim       · lim cos θ = 2 · 1 · 1 = 1
θ→0    θ     θ→0       θ            θ→0 θ       θ→0



Solution (ii)
Use a change of variable. Let ϕ = 2θ. Then
                sin 2θ         sin 2θ         sin ϕ
             lim       = 2 lim        = 2 lim       =1
            θ→0    θ       θ→0 2θ        ϕ→0 ϕ

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Lesson 10: Derivatives of Trigonometric Functions

  • 1. Section 3.4 Derivatives of Trigonometric Functions Math 1a February 25, 2008 Announcements Get 50% of all ALEKS points between now and 3/7 Problem Sessions Sunday, Thursday, 7pm, SC 310 Office hours Tuesday, Wednesday 2–4pm SC 323 Midterm I Friday 2/29 in class (up to §3.2)
  • 2. Two important trigonometric limits Theorem The following two limits hold: sin θ lim =1 θ→0 θ cos θ − 1 lim =0 θ→0 θ
  • 3. Proof of the Sine Limit Proof. Notice sin θ ≤ θ ≤ tan θ Divide by sin θ: θ 1 1≤ ≤ sin θ cos θ Take reciprocals: sin θ θ tan θ θ sin θ 1≥ ≥ cos θ cos θ 1 θ As θ → 0, the left and right sides tend to 1. So, then, must the middle expression.
  • 4. Now 1 − cos θ 1 − cos θ 1 + cos θ 1 − cos2 θ = · = θ θ 1 + cos θ θ(1 + cos θ) 2 sin θ sin θ cos θ = = · θ(1 + cos θ) θ 1 + cos θ So 1 − cos θ sin θ cos θ lim = lim · lim θ→0 θ θ→0 θ θ→0 1 + cos θ = 1 · 0 = 0.
  • 5. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx Proof. From the definition:
  • 6. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h
  • 7. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h (sin x cos h + cos x sin h) − sin x = lim h→0 h
  • 8. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h (sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h
  • 9. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx Proof. From the definition: d sin(x + h) − sin x sin x = lim dx h→0 h (sin x cos h + cos x sin h) − sin x = lim h→0 h cos h − 1 sin h = sin x · lim + cos x · lim h→0 h h→0 h = sin x · 0 + cos x · 1 = cos x
  • 10. Illustration of Sine and Cosine y x π π 0 π π − 2 2 sin x
  • 11. Illustration of Sine and Cosine y x π π 0 π π − cos x 2 2 sin x
  • 12. Derivatives of Sine and Cosine Theorem d sin x = cos x. dx d cos x = − sin x. dx
  • 13. Derivatives of tangent and secant Fill in the table: y y sin x cos x cos x − sin x tan x cot x sec x csc x
  • 14. Derivatives of tangent and secant Fill in the table: y y sin x cos x cos x − sin x tan x sec2 x cot x sec x csc x
  • 15. Derivatives of tangent and secant Fill in the table: y y sin x cos x cos x − sin x tan x sec2 x cot x − csc2 x sec x csc x
  • 16. Derivatives of tangent and secant Fill in the table: y y sin x cos x cos x − sin x tan x sec2 x cot x − csc2 x sec x sec x tan x csc x
  • 17. Derivatives of tangent and secant Fill in the table: y y sin x cos x cos x − sin x tan x sec2 x cot x − csc2 x sec x sec x tan x csc x − csc x cot x
  • 18. Example Find sin 2θ lim θ→0 θ
  • 19. Example Find sin 2θ lim θ→0 θ Solution (i) Use a trig identity: sin 2θ 2 sin θ cos θ sin θ lim = lim = 2 · lim · lim cos θ = 2 · 1 · 1 = 1 θ→0 θ θ→0 θ θ→0 θ θ→0
  • 20. Example Find sin 2θ lim θ→0 θ Solution (i) Use a trig identity: sin 2θ 2 sin θ cos θ sin θ lim = lim = 2 · lim · lim cos θ = 2 · 1 · 1 = 1 θ→0 θ θ→0 θ θ→0 θ θ→0 Solution (ii) Use a change of variable. Let ϕ = 2θ. Then sin 2θ sin 2θ sin ϕ lim = 2 lim = 2 lim =1 θ→0 θ θ→0 2θ ϕ→0 ϕ