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Section 0.4 Inverse Trigonometric Func- Tions

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Section 0.4 Inverse Trigonometric Func- Tions 1 Section 0.4 Inverse Trigonometric Func- tions Short Recall from Trigonometry De¯nition: A function f is periodic of period T if f(x + T ) = f(x) for all x such that x and x+T are in the domain of f. The smallest such number T > 0 is called the fundamental period. Example y = sin x is a periodic function with fundamental period (or just period) 2¼. sin(x + 2¼) = sin x cos 2¼ + cos x sin 2¼ = sin x The graph of y = sin x below is obtained by plotting points for 0 · x · 2¼ then using the periodic nature of the function to complete the graph. Note that the domain of y = sin x is (¡1; 1) i.e. all reals and range of f is the closed interval [¡1; 1]. (Also recall sin x = 0 if x = n¼ where n is an integer) 1 Out[4]= Π Π Π Π -2 Π - 3 -Π - Π 3 2 Π 2 2 2 2 -1 Figure 1: y=sin x 2 Example y = cos x is a periodic function with fundamental period (or just period) 2¼. cos(x + 2¼) = cos x cos 2¼ ¡ sin x sin 2¼ = cos x The graph of y = cos x below is also obtained by plotting points for one ¡¼ ¼ period (could be for 0 · x · 2¼ or 2 · x · 2 ) then using the periodic nature of the function to complete the graph. Note that the domain of y = cos x is also (¡1; 1) i.e all reals and range of f is the closed interval [¡1; 1]. (Also recall cos x = 0 if x = n¼=2 where n is an odd integer) 1 Out[5]= Π Π Π Π -2 Π - 3 -Π - Π 3 2 Π 2 2 2 2 -1 Figure 2: y=cos x The graph of the remaining four trig functions are shown at the end of these lecture notes and their domains can be written directly looking at their graphs. Notice also that tangent and cotangent have range (¡1; 1), where as cosecant and secant have range (¡1; ¡1] [ [1; 1). All four functions are periodic: tangent and cotangent have period ¼ whereas cosecant and secant have period 2¼. Inverse Trigonometric Functions When we try to ¯nd the the inverse of the trig. functions we have slight di±culty they are not 1-1. The solution out of this problem is to restrict the domain of these functions so that they become one to one. For an example apply the horizontal line test to the graph of y = sin x above. But the function f(x) = y = sin x for ¡¼=2 · x · ¼=2 is 1-1. 3 1 - Π Π 2 2 -1 The inverse function of this restricted sine function f exits and is denoted by sin¡1 or arcsin. It is called the inverse sine function or the arcsine function. Using what we learnt about graphing the inverse functions we can obtain the graph of sin¡1 x, reflect the graph with respect to the y = x line 1 - Π Π 2 2 -1 to get 4 Out[7]= -1 1 Note that sin x :[¡¼=2; ¼=2] ! [¡1; 1] and arcsin x :[¡1; 1] ! [¡¼=2; ¼=2]. Since the de¯nition of an inverse function says that f ¡1(x) = y , f(y) = x so we have ¡¼ ¼ sin¡1 x = y , sin y = x and · y · 2 2 or sin(sin¡1 x) = x for x 2 [¡1; 1] and sin¡1(sin x) = x for x 2 [¡¼=2; ¼=2] Warning Example sin¡1(sin ¼) = sin¡1(0) = 0. ¼ is not in the domain of arcsin so not ending up with ¼ is not a surprise. p ¡1 3 Example Evaluate sin ( 2 ) p ¡1 3 By the de¯nition of sin we want an angle θ 2 [¡¼=2; ¼=2] where sin θ = 2 ¼ so θ = 3 . ¡1 ¡1 Example Evaluate sin ( 2 ) ¡1 ¡¼ Find θ s.t sin θ = 2 so θ = 6 ¡1 1 Example Find the domain of f(x) for f(x) = sin ( x ). Domain of arcsin is [¡1; 1] so we are interested in those real numbers x such 1 that ¡1 · x · 1. And this inequality is satis¯ed if x 2 (¡1; ¡1] [ [1; 1). So Domain of f=(¡1; ¡1] [ [1; 1) Example Find the inverse of y = cos x. Cosine is, like sine, not 1-1 so we cannot ¯nd the inverse of it unless we restrict our domain. We will restrict our domain to be [0; ¼]. As you can check from the graph of cos x is 5 1-1 on this interval hence invertible. The inverse cosine function cos¡1 or arccos is de¯ned as: y = cos¡1 x i® cos y = x and 0 · y · ¼ So cos¡1 x has domain [¡1; 1] and range [0; ¼]. The below graph shows the process of reflecting the graph of y = cos x about the y = x line. 1 Out[6]= Π Π 2 -1 so the graph of y = arccos x is 2 Out[3]= - 1 p1 ¡ 2 Example Evaluate arccos( 2 ) p ¡ 2 We need to ¯nd the angle 0 · θ · ¼ where cos θ = 2 . Since cosine of our angle is negativep we expect our angle to be in the 2nd quadrant hence ¡ 2 θ = 3¼=4 or arccos( 2 ) = 3¼=4. Example Find the inverse of y = tan x. As in the cases of sine and cosine we need to restrict the domain of tangent to ¯nd its inverse. And we will restrict the domain of tan x to the interval (¡¼=2; ¼=2) where it is 1-1. Range of tan x on this domain is all <. The inverse of tangent function tan¡1 or arctan is de¯ned just like in the cases of sine and cosine. Domain 6 of arctan x is all < and range is (¡¼=2; ¼=2). And the graph of y = arctan x is: Π 2 Out[5]= - Π 2 Remark: Note that this is an example of a function with two di®erent hor- izontal asymptotes. Think of the horizontal asymptotes for the time being the horizontal lines with which graph of y = arctan x becomes "snug" in the long run (as x becoming larger and larger or smaller and smaller.). Judging our graph based on this "rough" de¯nition you can say that arctan x be- comes "snug" with y = ¡¼=2 line as x becomes smaller and smaller and it becomes "snug" with the y = ¼=2 line as x becomes larger and larger. Example Simplify the expression cos(tan¡1(x)) ¡1 Let y = tan x then x = tan y where ¡¼=2 <p y < ¼=2. Construct the tri- angle with angle y, and whose hypotenuse is 1 + x2 and x = tan y. From here you can easily read cos(tan¡1(x)) = cos y = p 1 1+x2 Rest of the inverses y = cot¡1 x(x 2 <) , cot y = x and y 2 (0; ¼) y = sec¡1 x(jxj ¸ 1) , sec y = x and y 2 [0; ¼=2) [ [¼; 3¼=2) y = csc¡1 x(jxj ¸ 1) , csc y = x and y 2 [0; ¼=2) [ [¼; 3¼=2) Last graph I'll provide will be for sec¡1 x below you should try out the other two yourself. 7 Π Π Out[9]= 2 -1 1 Pictorial hint for ¯nding the inverse: The way to ¯nd the graph of the inverse function is to rotate your paper (which has the graph on it) by ¼ degrees around the main diagonal (the line through the origin at angle ¼=4 counterclockwise from the x axis.) You will then ¯nd that you have to look through your paper at the function but that can usually be done and if you start with the graph of f you are looking at the graph of the inverse function to f. 8 1 Out[6]= 3 Π Π Π 3 Π -2 Π - -Π - - Π 2 Π 2 2 1 2 2 Figure 3: y=tan x 1 Out[8]= 3 Π Π Π 3 Π -2 Π - -Π - - Π 2 Π 2 2 1 2 2 Figure 4: y=cot x 1 Out[9]= 3 Π Π Π 3 Π -2 Π - -Π - -1 Π 2 Π 2 2 2 2 Figure 5: y=sec x 1 Out[10]= Π Π Π Π -2 Π - 3 -Π - -1 Π 3 2 Π 2 2 2 2 Figure 6: y=csc x.
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