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Derivatives of Trigonometric Functions Chapter 3.5 Proving that sin π₯ lim π₯β0 π₯ =1 β’ In section 2.1 you used a table of values approaching 0 from the left sin π₯ and right that lim = 1; but that was not a proof π₯β0 π₯ β’ Because you will need to know this limit (and a related one for cosine), we will begin this section by proving this through geometry β’ We will need the following β’ The Squeeze Theorem β’ The formula for the area of a sector of a circle 2 Proving that sin π₯ lim π₯β0 π₯ =1 β’ The Squeeze Theorem says that if π, π, and β are functions so that, for π₯-values in an open interval around some number π if π π₯ β€ π π₯ β€ β π₯ and lim π(π₯) = lim β(π₯) = πΏ π₯βπ π₯βπ then lim π(π₯) = πΏ π₯βπ 3 Proving that sin π₯ lim π₯β0 π₯ =1 β’ If a circle of radius 1 is subtended by an angle π in radians (creating a sector or βpie sliceβ), then the area of the sector is π π 2 ππ β = 2π 2 4 Proving that sin π₯ lim π₯β0 π₯ =1 5 Proving that sin π₯ lim π₯β0 π₯ =1 sin π π tan π β€ β€ 2 2 2 β’ We can simplify by multiplying through by 2; next divide through by sin π (we can do this without changing inequality signs; why?) π 1 1β€ β€ sin π cos π 1 1 β’ For positive numbers π₯ and π¦, if π₯ β₯ π¦ then β€ π₯ 1 3 β’ For example, 3 β₯ 2, but β€ 1 ; 2 π¦ we use this to take the reciprocals 6 Proving that sin π₯ lim π₯β0 π₯ =1 π 1 1β€ β€ sin π cos π sin π 1β₯ β₯ cos π π sin π cos π β€ β€1 π sin π β’ This brings in and also puts it between two functions within the π first quadrant; we are now able to use the Squeeze Theorem 7 Proving that Since sin π₯ lim π₯β0 π₯ =1 sin π cos π β€ β€1 π and since lim+ cos π = lim+ 1 = 1 πβ0 πβ0 then by the Squeeze Theorem sin π lim+ =1 πβ0 π 8 Proving that sin π₯ lim π₯β0 π₯ =1 Since sine is an odd function, then sin βπ β sin π sin π = = βπ βπ π We can conclude that sin π sin π limβ = lim+ =1 πβ0 πβ0 π π Hence, sin π lim =1 πβ0 π QED 9 Proving that 1βcos π₯ lim π₯ π₯β0 =0 We can use the previous result to prove another special limit, specifically 1 β cos π₯ lim =0 π₯β0 π₯ We have 1 β cos π₯ 1 + cos π₯ 1 β cos2 π₯ lim β = lim π₯β0 π₯ 1 + cos π₯ π₯β0 π₯ 1 + cos π₯ We want to be able to use the previous theorem. Note that the numerator is just sin2 π₯ 10 Proving that 1βcos π₯ lim π₯ π₯β0 =0 1 β cos π₯ 1 + cos π₯ 1 β cos2 π₯ lim β = lim π₯β0 π₯ 1 + cos π₯ π₯β0 π₯ 1 + cos π₯ sin2 π₯ = lim π₯β0 π₯ 1 + cos π₯ Now use the limit properties to get sin π₯ sin π₯ 0 = lim β lim =1β =0 π₯β0 π₯ π₯β0 1 + cos π₯ 1+1 11 Derivative of the Sine and Cosine Functions sin π₯ lim π₯β0 π₯ 1βcos π₯ lim π₯ π₯β0 β’ Having proved that = 1 and that = 0, we can now use the derivative definition to determine the derivative of the sine function and the cosine function β’ We will also need to be reminded of the following trigonometric identity: sin π΄ + π΅ = sin π΄ cos π΅ + sin π΅ cos π΄ 12 Derivative of the Sine Function THEOREM: If π π₯ = sin π₯, then π β² π₯ = cos π₯ 13 Derivative of the Sine Function THEOREM: PROOF Using the definition of the derivative: sin π₯ + β β sin π₯ π π₯ = lim ββ0 β Use the previous identity to expand the numerator: β² sin π₯ cos β + sin β cos π₯ β sin π₯ = lim ββ0 β 14 Derivative of the Sine Function THEOREM: PROOF sin π₯ cos β + sin β cos π₯ β sin π₯ = lim ββ0 β We can factor sin π₯ sin π₯ cos β β 1 + sin β cos π₯ = lim ββ0 β Now use the limit properties 15 Derivative of the Sine Function THEOREM: PROOF sin π₯ cos β β 1 + sin β cos π₯ = lim ββ0 β β 1 β cos β sin β = lim sin π₯ β lim + lim cos π₯ β lim ββ0 ββ0 ββ0 ββ0 β β = sin π₯ β 0 + cos π₯ β 1 = cos π₯ QED 16 Derivative of the Cosine Function THEOREM: If π π₯ = cos π₯, then π β² π₯ = β sin π₯ 17 Derivative of the Cosine Function THEOREM: PROOF We will need the angle sum identity for cosine which is cos(π΄ + π΅) = cos π΄ cos π΅ β sin π΄ sin π΅ Use the definition of the derivative: πβ² cos π₯ + β β cos π₯ π₯ = lim ββ0 β Expand the numerator: 18 Derivative of the Cosine Function THEOREM: PROOF πβ² cos π₯ + β β cos π₯ π₯ = lim ββ0 β cos π₯ cos β β sin π₯ sin β β cos π₯ = lim ββ0 β Factor out cos π₯ cos π₯ cos β β 1 β sin π₯ sin β = lim ββ0 β 19 Derivative of the Cosine Function THEOREM: PROOF cos π₯ cos β β 1 β sin π₯ sin β = lim ββ0 β Use the limit properties β cos β β 1 sin β = lim cos π₯ β lim β lim sin π₯ β lim ββ0 ββ0 ββ0 ββ0 β β = cos π₯ β 0 β sin π₯ β 1 = β sin π₯ QED 20 Example 1: Using the Derivatives of Sine and Cosine Functions Find the derivatives of a) π¦ = π₯ 2 sin π₯ b) π’ = cos π₯ 1βsin π₯ 21 Example 1: Using the Derivatives of Sine and Cosine Functions Find the derivatives of a) π¦ = π₯ 2 sin π₯ Use the Product Rule: ππ¦ π π 2 2 =π₯ β sin π₯ + sin π₯ β π₯ ππ₯ ππ₯ ππ₯ = π₯ 2 cos π₯ + 2π₯ sin π₯ 22 Example 1: Using the Derivatives of Sine and Cosine Functions Find the derivatives of cos π₯ b) π’ = 1βsin π₯ Use the Quotient Rule ππ’ = ππ₯ 1 β sin π₯ β π π cos π₯ β cos π₯ β 1 β sin π₯ ππ₯ ππ₯ 1 β sin π₯ 2 β sin π₯ 1 β sin π₯ β cos π₯ β β cos π₯ β sin π₯ + sin2 π₯ + cos 2 π₯ = = 1 β sin π₯ 2 1 β sin π₯ 2 1 β sin π₯ 1 = = 2 1 β sin π₯ 1 β sin π₯ 23 Simple Harmonic Motion 24 Example 2: The Motion of a Weight on a Spring A weight hanging from a spring is stretched 5 units beyond its rest position (π = 0) and released at time π‘ = 0 to bob up and down. Its position at any later time π‘ is π π‘ = 5 cos π‘ What are its velocity and acceleration at time π‘? Describe its motion. 25 Example 2: The Motion of a Weight on a Spring A weight hanging from a spring is stretched 5 units beyond its rest position (π = 0) and released at time π‘ = 0 to bob up and down. Its position at any later time π‘ is π π‘ = 5 cos π‘ What are its velocity and acceleration at time π‘? Describe its motion. Recall that velocity is the first derivative of position and acceleration is the second derivative of position (or the first derivative of velocity). π£ π‘ = π β² π‘ = β5 sin π‘ π π‘ = π£ β² π‘ = π β²β² π‘ = β5 cos π‘ 26 Example 2: The Motion of a Weight on a Spring Time π π = π cos π π π = βπ sin π π π = βπ cos π 0 π 2 π 3π 2 2π 5π 2 3π 7π 2 4π 5 0 β5 0 β5 0 β5 0 5 0 5 0 5 0 β5 0 β5 0 β5 0 5 0 5 0 5 0 β5 27 Jerk β’ A sudden change in acceleration is called a jerk β’ When a ride in a vehicle is jerky, it is caused by sudden changes in acceleration β’ Hence, jerk, sudden changes in acceleration, is the first derivative of acceleration, or the third derivative of acceleration 28 Example 3: A Couple of Jerks a) What is the jerk caused by the acceleration due to gravity? b) What is the jerk of the simple harmonic motion from Example 2? 29 Example 3: A Couple of Jerks a) What is the jerk caused by the acceleration due to gravity? π2 π ππ‘ 2 The acceleration due to gravity is constant (near the Earthβs surface): = β32 feet per second per second. Since the derivative of a constant is zero, then π3π =0 3 ππ‘ b) What is the jerk of the simple harmonic motion from Example 2? ππ = 5 sin π‘ ππ‘ Jerk is maximum when sin π‘ = ±1, which occurs when the weight is at the rest position and acceleration changes sign 30 Derivatives of the Other Basic Trigonometric Functions β’ We can use the derivatives of sine and cosine as well as the Product and/or Quotient Rules to determine the derivatives of the other basic trigonometric functions β’ Find π [tan π₯] ππ₯ π π sin π₯ tan π₯ = ππ₯ ππ₯ cos π₯ cos π₯ β cos π₯ β sin π₯ β β sin π₯ 1 2 = = = sec π₯ 2 2 cos π₯ cos π₯ 31 Derivatives of the Other Basic Trigonometric Functions β’ Find β’ π ππ₯ cot π₯ β’ π ππ₯ sec π₯ β’ π [csc π₯] ππ₯ 32 Derivatives of the Other Basic Trigonometric Functions THEOREM: β’ π ππ₯ tan π₯ = sec 2 π₯ β’ π ππ₯ cot π₯ = β csc 2 π₯ β’ π ππ₯ sec π₯ = tan π₯ sec π₯ β’ π ππ₯ csc π₯ = β cot π₯ csc π₯ 33 Example 4: Finding Tangent and Normal Lines Find equations for the lines that are tangent and normal to the graph of tan π₯ π π₯ = π₯ at π₯ = 2. 34 Example 4: Finding Tangent and Normal Lines Find equations for the lines that are tangent and normal to the graph of tan π₯ π π₯ = π₯ at π₯ = 2. 2 π₯ β tan π₯ β 1 2 π₯ β tan π₯ π₯ β sec π₯ sec πβ² π₯ = = 2 π₯ π₯2 The slope of the tangent line at π₯ = 2 is π β² 2 β 3.433 and of the normal line is 1 β β² β β0.291. The equations are π¦ β 3.433 π₯ β 2 β 1.093 and π¦ β π 2 β 0.291 π₯ β 2 β 1.093 35 Example 5: A Trigonometric Second Derivative Find π¦β²β² if π¦ = sec π₯. 36 Example 5: A Trigonometric Second Derivative Find π¦β²β² if π¦ = sec π₯. π¦ β² = tan π₯ sec π₯ π¦ β²β² = tan π₯ β tan π₯ sec π₯ + sec π₯ β sec 2 π₯ = tan2 π₯ sec π₯ + sec 3 π₯ 37 Exercise 3.5 38