Math 103 Lecture 6 Dr. Mustafa El-Agamy Summary Scaling Property π = π π¬π’π§ π π = π¬π’π§ π π = π. π π¬π’π§ π The three functions have the same properties except the range. π = π¬π’π§ π π = π¬π’π§(π π) Period= π π = π¬π’π§ π π = π¬π’π§(π. π π) Period= π π The three functions have the same properties except the period and the zeros. Scaling Property Function π = π¨ π¬π’π§(π π) π = π¨ ππ¨π¬(π π) π = π¨ πππ§(π π) Domain β β β− ππ 2π π: odd integer Range Period Zeros −π¨, π¨ 2π π ππ π −π¨, π¨ 2π π ππ 2π β π π ππ π π¨>π π: integer π: odd integer π: integer Sketch the function π = π π¬π’π§ ππ. Then, state the domain, range, period zeros and symmetry. Domain: −∞, ∞ ≡ β Range: −2, 2 Period: 2π/3 Zeros: π π/3 π: integer Symmetry: Odd function Sketch the function π = π π¬π’π§ π . Then, state the domain, range, period zeros and symmetry. Domain: −∞, ∞ ≡ β Range: 0, 3 Period: π Zeros: π π π: integer Symmetry: Even function Find the range of the function π = π − π ππ¨π¬ ππ. Also, find the period and the zeros. (Multiply by -3) ⇒ β΅ −1 ≤ cos 4π₯ ≤ 1 (Add 8) ⇒ 11 ≥ 8 − 3 cos 4π₯ ≥ 5 Period: 2π 4 = π 2 3 ≥ −3 cos 4π₯ ≥ −3 ⇒ 11 ≥ π¦ ≥ 5 Zeros: ∴ π¦ ∈ 5 , 11 No π₯ − intercepts Trigonometric and Hyperbolic Functions Trigonometric Identities cos(π΄ − π΅) = cos π΄ . cos π΅ + sin π΄ . sin π΅ (1) cos(π΄ + π΅) = cos π΄ . cos π΅ − sin π΄ . sin π΅ (2) sin(π΄ − π΅) = sin π΄ . cos π΅ − cos π΄ . sin π΅ (3) sin(π΄ + π΅) = sin π΄ . cos π΅ + cos π΄ . sin π΅ (4) If π΄ = π΅ = π₯ , we get from (1) cos 2 π₯ + sin2 π₯ = 1 ÷ cos 2 π₯ ÷ sin2 π₯ 1 + tan2 π₯ = sec 2 π₯ cot 2 π₯ + 1 = csc 2 π₯ from (2) cos(2π₯) = cos 2 π₯ − sin2 π₯ cos(2π₯ ) = 2 cos 2 π₯ − 1 cos(2π₯ ) = 1 − 2 sin2 π₯ 1 cos 2 π₯ = (1 + cos 2π₯) 2 1 sin2 π₯ = (1 − cos 2π₯) 2 from (4) sin(2π₯) = 2 sin π₯ . cos π₯ 8 Trigonometric and Hyperbolic Functions Prove that 2 cos 7π₯ . cos 3π₯ = cos 4π₯ + cos 10π₯ Solution By adding equation (1) and (2), we get cos(π΄ − π΅) + cos(π΄ + π΅) = 2 cos π΄ . cos π΅ Then put π΄ = 7π₯ and π΅ = 3π₯. Prove that tan(π΄ + π΅) = tan π΄ + tan π΅ 1 − tan π΄ . tan π΅ Solution tan(π΄ + π΅) = sin(π΄ + π΅) sin π΄ . cos π΅ + cos π΄ . sin π΅ = cos(π΄ + π΅) cos π΄ . cos π΅ − sin π΄ . sin π΅ ÷ cos π΄ . cos π΅ sin π΄ . cos π΅ cos π΄ . sin π΅ (cos π΄ . cos π΅) + (cos π΄ . cos π΅) tan π΄ + tan π΅ = = sin π΄ . sin π΅ 1 − tan π΄ . tan π΅ 1−( cos π΄ . cos π΅) Note that: Replace π΅ by (−π΅) For π΄ = π΅ = π₯ tan(2π₯) = 2 tan π₯ 1 − tan2 π₯ tan(π΄ − π΅) = 9 tan π΄ − tan π΅ 1 + tan π΄ . tan π΅ Trigonometric and Hyperbolic Functions Geometry Angle between two lines (π) πΌ1 = π + πΌ2 π = πΌ1 − πΌ2 π tan π = tan(πΌ1 − πΌ2 ) πΆπ tan πΌ1 − tan πΌ2 tan π = 1 + tan πΌ1 . tan πΌ2 ∴ tan π = πΆπ π1 − π2 1 + π1 . π2 where π = πππ§ πΆ is the line slope. Parallel lines (π = 0° or 180°): 0= π1 − π2 ⇒ π1 − π2 = 0 ⇒ 1 + π1 . π2 π1 = π2 Perpendicular lines (π = 90°): 1 π1 − π2 = ⇒ 1 + π1 . π2 = 0 ⇒ 0 1 + π1 . π2 10 π1 . π2 = −1 π Trigonometric and Hyperbolic Functions Radius of Earth One thousand two hundred years ago, the great scientist Abu Al-Rayhan Al-Biruni invented an easy calculation method to measure the radius of Earth by means of Astrolabe. Astrolabe is an instrument used to measure the angles of elevation and depression. 1 He measured the height of a mountain π above sea level using trigonometry. He measured an elevation angle π½π , then moved toward the mountain a distance π , then measured another elevation angle π½π . tan π1 = β π+β tan π2 = β β π΅ tan π1 β = π+( β tan π2 ) tan π1 = β tan π2 π tan π2 + β 11 β= π tan π1 tan π2 tan π2 − tan π1 Trigonometric and Hyperbolic Functions 2 Al-Biruni ascended on the mountain that knows its height π and set the pointer of Astrolabe so that it indicated the last point in the sea C. So that, he could measure the angle of depression πΆ. The tangent AC (Line of sight) is perpendicular to the radius of the circle πΆ π (Earth) cos πΌ = π π +β πΉ πΉ π cos πΌ + β cos πΌ = π πΆ πΆ π = β cos πΌ 1 − cos πΌ Al-Biruni repeated the measurement several times and averaged the results to increase the accuracy of the measurement. Al-Biruni was able to measure only four quantities (three angles and a distance) to calculate the radius of Earth with an accuracy 99%. ο It is truly amazing how mathematics enables a man sitting on top of a mountain to measure the size of the world without traveling from where he is. Did you know the importance of mathematics now?! 12 Trigonometric and Hyperbolic Functions Hyperbolic Functions Certain even and odd combinations of the exponential functions ππ and π−π arise so frequently in mathematics and its applications. In many ways they are analogous to the trigonometric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason, they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic cosine, and so on. Hyperbola The graphs of hyperbolic sine and cosine π π₯ − π −π₯ π¦ = sinh π₯ = 2 π π₯ + π −π₯ π¦ = cosh π₯ = 2 can be sketched using graphical addition as in the following figures. 13 Trigonometric and Hyperbolic Functions Domain: π₯ ∈ β Domain: π₯ ∈ β Range: π¦ ∈ β Range: π¦ ∈ [1, ∞[ Odd function Even function sinh(−π₯) = − sinh π₯ cosh(−π₯) = cosh π₯ Zero: π₯ = { 0 } sinh π₯ π π₯ − π −π₯ π 2π₯ − 1 1 − π −2π₯ π¦ = tanh π₯ = = = = cosh π₯ π π₯ + π −π₯ π 2π₯ + 1 1 + π −2π₯ Domain: π₯ ∈ β Range: π¦ ∈ ]−1,1[ Zero: π₯ = { 0 } Odd function tanh(−π₯) = − tanh π₯ lim tanh π₯ = 1 π₯→∞ lim tanh π₯ = −1 π₯ → −∞ Horiz. asymp.: π¦ = ±1 14 Trigonometric and Hyperbolic Functions 1 2 2 ππ₯ π¦ = csch π₯ = = = sinh π₯ π π₯ − π −π₯ π 2π₯ − 1 Domain: π₯ ∈ β − { 0 } Range: π¦ ∈ β − { 0 } Odd function csch(−π₯) = − csch π₯ lim csch π₯ = −∞ π₯ → 0− lim csch π₯ = ∞ π₯ → 0+ lim csch π₯ = 0 π₯ → −∞ lim csch π₯ = 0 π₯→∞ Asymptotes: Vertical (π₯ = 0) and Horizontal ( π¦ = 0) 1 2 2 ππ₯ π¦ = sech π₯ = = = cosh π₯ π π₯ + π −π₯ π 2π₯ + 1 Domain: π₯ ∈ β Range: π¦ ∈ ]0,1] Even function sech(−π₯) = sech π₯ lim sech π₯ = 0 π₯→∞ lim sech π₯ = 0 π₯ → −∞ Horizontal asymp.: π¦ = 0 15 Trigonometric and Hyperbolic Functions 1 cosh π₯ π π₯ + π −π₯ π 2π₯ + 1 1 + π −2π₯ π¦ = coth π₯ = = = = = tanh π₯ sinh π₯ π π₯ − π −π₯ π 2π₯ − 1 1 − π −2π₯ Domain: π₯ ∈ β − { 0 } Range: π¦ ∈ β − [−1,1] Odd function coth(−π₯) = − coth π₯ lim coth π₯ = −∞ lim coth π₯ = ∞ π₯ → 0− π₯ → 0+ lim coth π₯ = −1 lim coth π₯ = 1 π₯ → −∞ π₯→∞ Asymptotes: Vertical (π₯ = 0) and Horizontal ( π¦ = ±1) 16 Trigonometric and Hyperbolic Functions Prove that cosh π₯ + sinh π₯ = π π₯ Solution π π₯ + π −π₯ π π₯ − π −π₯ 2 π π₯ LHS = cosh π₯ + sinh π₯ = + = = π π₯ = RHS 2 2 2 Note that: cosh π₯ + sinh π₯ = π π₯ cosh π₯ − sinh π₯ = π −π₯ Prove that (cosh π₯ + sinh π₯)5 = cosh(5π₯) + sinh(5π₯) Solution LHS = (cosh π₯ + sinh π₯)5 = (π π₯ )5 = π 5π₯ = cosh(5π₯) + sinh(5π₯) = RHS Note that: (cosh π₯ + sinh π₯)π = cosh(π π₯) + sinh(π π₯) (cosh π₯ − sinh π₯)π = cosh(π π₯) − sinh(π π₯) for any π ∈ β 17 Trigonometric and Hyperbolic Functions Hyperbolic Identities cosh(π΄ − π΅) = cosh π΄ . cosh π΅ − sinh π΄ . sinh π΅ (5) cosh(π΄ + π΅) = cosh π΄ . cosh π΅ + sinh π΄ . sinh π΅ (6) sinh(π΄ − π΅) = sinh π΄ . cosh π΅ − cosh π΄ . sinh π΅ (7) sinh(π΄ + π΅) = sinh π΄ . cosh π΅ + cosh π΄ . sinh π΅ (8) If π΄ = π΅ = π₯ , we get from (5) cosh2 π₯ − sinh2 π₯ = 1 ÷ cosh2 π₯ ÷ sinh2 π₯ 1 − tanh2 π₯ = sech2 π₯ coth2 π₯ − 1 = csch2 π₯ from (6) cosh(2π₯) = cosh2 π₯ + sinh2 π₯ cosh(2π₯) = 2 cosh2 π₯ − 1 cosh(2π₯) = 2 sinh2 π₯ + 1 1 cosh2 π₯ = (cosh 2π₯ + 1) 2 1 sinh2 π₯ = (cosh 2π₯ − 1) 2 from (8) sinh(2π₯) = 2 sinh π₯ . cosh π₯ 18 Trigonometric and Hyperbolic Functions Note that: (cosh π₯ + sinh π₯)π = cosh(π π₯) + sinh(π π₯) For π = 2 : LHS = (cosh π₯ + sinh π₯)2 = cosh2 π₯ + sinh2 π₯ + 2 sinh π₯ . cosh π₯ = cosh(2π₯) + sinh(2π₯) = RHS Prove that sinh(2π₯) = 2 sinh π₯ . cosh π₯ Solution RHS = 2 sinh π₯ . cosh π₯ π π₯ − π −π₯ π π₯ + π −π₯ π 2π₯ − π −2π₯ = 2( = sinh(2π₯) = LHS ).( )= 2 2 2 Prove that cosh2 π₯ − sinh2 π₯ = 1 Solution π π₯ + π −π₯ 2 π π₯ − π −π₯ 2 LHS = cosh π₯ − sinh π₯ = ( ) −( ) 2 2 2 2 π 2π₯ + π −2π₯ + 2 π 2π₯ + π −2π₯ − 2 4 =( )−( ) = = 1 = RHS 4 4 4 19 Trigonometric and Hyperbolic Functions Another solution technique LHS = cosh2 π₯ − sinh2 π₯ = (cosh π₯ + sinh π₯). (cosh π₯ − sinh π₯) = π π₯ . π −π₯ = 1 = RHS Prove that cosh(π΄ + π΅) = cosh π΄ . cosh π΅ + sinh π΄ . sinh π΅ Solution RHS = cosh π΄ . cosh π΅ + sinh π΄ . sinh π΅ π π΄ + π −π΄ π π΅ + π −π΅ π π΄ − π −π΄ π π΅ − π −π΅ =( ).( )+( ).( ) 2 2 2 2 π π΄+π΅ + π π΄−π΅ + π −π΄+π΅ + π −π΄−π΅ π π΄+π΅ − π π΄−π΅ − π −π΄+π΅ + π −π΄−π΅ =( )+( ) 4 4 2π π΄+π΅ + 2π −π΄−π΅ π (π΄+π΅) + π −(π΄+π΅) = = = cosh(π΄ + π΅) = LHS 4 2 Prove that 1 + tanh π₯ ππ₯ = √ 1 − tanh π₯ Solution sinh π₯ 1 + tanh π₯ 1 + (cosh π₯ ) cosh π₯ + sinh π₯ ππ₯ = = = −π₯ = π 2π₯ sinh π₯ 1 − tanh π₯ 1 − cosh π₯ − sinh π₯ π ( ) cosh π₯ 1 + tanh π₯ RHS = √ = √π 2π₯ = π π₯ = LHS 1 − tanh π₯ 20