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Section 2.11 Notes Page 1 2.11 Inverse Trigonometric Functions 1 . We put in an angle and get a value as a result. In inverse trig 2 1 functions we put in the value and get an angle: sin 1 30 . So here we put in the value of one half and got 2 30 degrees as a result. We are not allowed to put any number into our inverse trig functions. There are restrictions on the domain that are given in the following table: From trigonometry we know that sin 30 1 y sin x Domain 1 x 1 y cos x 1 x 1 y tan1 x x 1 y cot x x y sec1 x x 1 1 1 y csc x x 1 Range y 2 2 0 y y 2 2 0 y 0 y , y 2 y 2 2 , y0 We are not going to look at the graphs of the inverse trig functions in this class, however that is where the domain and range is derived from. 3 . EXAMPLE: Find sin 1 2 3 that has a value of .” You will need to 2 2 2 remember your unit circle or table for this one. This corresponds to an angle of 60 degrees, or , which is the 3 answer. What this is really asking is: “find an angle between and 1 EXAMPLE: Find cos 1 . 2 What this is really asking is: “find an angle between 0 and that has a value of 1 .” If you rationalize, then 2 1 2 . This corresponds to an angle of 45 degrees, or , which is the answer. 2 4 2 Section 2.11 Notes Page 2 3 . EXAMPLE: Find tan 1 3 3 that has a value of .” You will need to 2 3 2 remember your unit circle or table for this one. This corresponds to an angle of 30 degrees, or , which is the 6 answer. What this is really asking is: “find an angle between and EXAMPLE: Find: cot 1 ( 3 ) . What this is asking us to do is to find the angle that gives us a cotangent value of 0. Our answer needs to be between 0 and since that is our range for an inverse cotangent. Since cotangent is cosine divided by sine, we 5 can find a point on the unit circle such that the cosine over sine gives us a value of 3 . The answer is . 6 EXAMPLE: Find: csc1 ( 2 ) . What this is asking us to do is to find the angle that gives us a cosecant value of 0. Our answer needs to be between / 2 and / 2 since that is our range for an inverse cosecant. Since cosecant is 1 divided by sine, 7 we can find a point on the unit circle such that 1 divided by sine gives us a value of 2 . The answer is . 4 EXAMPLE: Find: sec1 (2) . What this is asking us to do is to find the angle that gives us a secant value of 2. Our answer needs to be between 0 and since that is our range for an inverse secant. Since secant is 1 divided by cosine, we can find a point on the unit circle such that 1 divided by cosine gives us a value of 2 . The answer is . 3 7 . EXAMPLE: Find: cos 1 2 7 as a decimal is 1.32. If we look on the table in this section, we see that the domain must be 2 7 is undefined. between -1 and 1. Since 1.32 falls outside of this, then this means cos 1 2 The number Section 2.11 Notes Page 3 EXAMPLE: Write in algebraic form: sectan 4 x . 1 These problems involve drawing a triangle and labeling the sides with algebraic expressions. For all these problems we will assume that x is positive and the triangle should be drawn in the first quadrant. We can 4x rewrite our problem as: sec tan 1 We know that the adjacent side is 1 and the opposite side is 4x. We can 1 use the Pythagorean theorem to find the hypotenuse: c 2 (4 x) 2 (1) 2 . So we have c 16 x 2 1 16 x 2 1 4x The secant on the outside of our problem tells us how to write our answer. From our drawing, secant is 16 x 2 1 over 1 so we write our answer as: 1 sec tan1 4 x 16 x 2 1 EXAMPLE: Write in algebraic form: cot sin 1 ( x 1) . These problems involve drawing a triangle and labeling the sides with algebraic expressions. For all these problems we will assume that x is positive and the triangle should be drawn in the first quadrant. We can x 1 rewrite our problem as: sec sin 1 We know that the opposite side is x – 1 and the hypotenuse is 1. We 1 can use the Pythagorean theorem to find the hypotenuse: 12 ( x 1) 2 a 2 . So we have 1 x 2 2 x 1 a 2 . Solving for a 2 we get a 2 2 x x 2 , so a 2 x x 2 . Now we can draw the triangle. 1 x–1 The cotangent on the outside of our problem tells us how to write our answer. From our drawing, cotangent is 2x x 2 so we write our answer as: cot sin 1 ( x 1) 2 x x 2 over x – 1 2x x2 where x 1 . x 1 Derivative of inverse trig functions Let’s find the derivative of sin 1 u . In order to do this we will draw a triangle like in the previous problems. u We can rewrite this problem as sin 1 . Then we know the hypotenuse is 1 and the opposite side is u. 1 Then we can use the Pythagorean Theorem to find the remaining side. You will get this triangle: 1 1 u2 u Section 2.11 Notes Page 4 On our problem, let’s let y sin u . The definition of the inverse sine tells us that sin y u . Now let’s take the derivative of both sides with respect to x using implicit differentiation: 1 d sin y du dx dx You will need to use the chain rule on the left side. cos y y u Now solve for y y y y u cos y From our triangle we know that cos y u Now simplify and we have our answer. 1 u2 1 u 1 u 1 u2 . Plug this in for cos y . 1 So if y sin 1 u then y 2 u 1 u2 . EXAMPLE: If y sec1 u , find y . In order to do this we will draw a triangle like in the previous problems. We can rewrite this problem as u sec 1 . Then we know the hypotenuse is u and the adjacent side is 1. 1 Then we can use the Pythagorean Theorem to find the remaining side. You will get this triangle: u 2 1 u 1 The definition of the inverse secant tells us that sec y u . Now let’s take the derivative of both sides with respect to x using implicit differentiation: d sec y du dx dx sec y tan y y u y y u sec y tan y u u u 1 2 You will need to use the chain rule on the left side. Now solve for y We know that sec y u and tan y u 2 1 from our triangle. Now substitute. So if y sec1 u then y u u u 2 1 . Section 2.11 Notes Page 5 Derivatives of Inverse Trig Functions d u sin 1 u dx 1 u2 d u cos 1 u dx 1 u2 d u tan 1 u dx 1 u2 d u cot 1 u dx 1 u2 d u csc1 u dx u u2 1 d u sec1 u dx u u2 1 EXAMPLE: Find the derivative: y sec1 (2 x) . We will let u 2 x . Then u 2 . We just need to substitute these into the formula y 2 2 x (2 x) 1 2 . Now we can simplify: y 1 x 4 x2 1 u u u 2 1 . You will get: . This is as far as we can go. EXAMPLE: Find the derivative: y tan1 5 2 x . 1 1 1 We will let u 5 2 x . Then u (5 x) 2 (2) . This simplifies to u . Now we need to 2 5 2x 1 u 5 2 x . Simplifying gives us: substitute these into the formula . You will get: y 2 1 u2 1 5 2x 1 1 1 1 1 y 5 2 x = 5 2 x = . So our final answer is: y . 6 2x 1 5 2x (6 2 x ) 5 2 x 5 2x 6 2x 1 EXAMPLE: Find the derivative: y x tan 1 (2 x) ln(1 4 x 2 ) . 4 We need to use the product rule on this one. When we take the derivative of tan1 (2 x) we will let u 2 x . u Then u 2 and we will use the formula the formula . For the second term since we have a natural log 1 u2 u we will let u 1 4 x 2 . Then u 8x . To do a derivative of the ln we will use the formula . Using the u 2 1 8x tan 1 (2 x) (1) product rule we get: y x . Now we can simplify: 2 1 ( 2 x) 4 1 4x2 2 2x y x tan 1 (2 x) . The first and last terms cancel and then we have: y tan1 (2 x) . 2 1 4x 1 4x2 1 Section 2.11 Notes Page 6 EXAMPLE: Find the derivative: y ln cot 2 x . 3 Since we have a natural log we will let u cot 1 4 x3 . Then u 6 x2 1 2 x3 2 . This simplifies: u 6x2 . To 1 4 x6 6x2 u 4 x 6 . Now we can take the derivative of the ln we will use the formula . This will give us: y 1 u cot 1 4 x3 6 x2 simplify: y 1 3 . Nothing more we can do on this one. cot 2 x 1 4 x6 x EXAMPLE: Find the derivative: y 25 sin 1 x 25 x 2 . Write your answer as a single fraction. 5 u x 1 . Then u . You will use . For the second term you will need to 5 5 1 u2 use the product rule combined with the chain rule. Putting this all together you will have: 1 1 1 5 y 25 x (25 x 2 ) 2 (2 x) 25 x 2 (1) Now we need to simplify. 2 2 x 1 5 For the first term we will let u y 5 1 y y y y y x2 25 5 25 x 25 2 5 25 x 5 x2 2 25 25 x 2 25 x 2 x2 25 x 25 x 2 25 x 2 . 2 x2 25 x 2 x2 25 x 2 25 x 2 (25 x 2 ) 2x 2 25 x 2 . For 1 x2 25 x 2 we can get common denominators: . 25 25 25 x 2 This simplified further. 25 x 2 The first term can be simplified further. 25 x 2 1 Multiply the last term by 25 x 2 25 x 2 to get common denominators. Now simplify the numerator and we are done.