Lecture 16: Inverse (§3.5)

Today: Inverse trigonometric functions, and their derivatives

Example 74 (Warm-Up). Compute f (x), where f(x)=(2x + x2)2x.

Last time, we learned that the way to compute derivatives of functions like this (expo- nential functions where both the base and the exponent are functions of x) is via logarithmic differentiation. Indeed, by first writing ln(f(x)) = ln((2x + x2)2x)=2x ln(2x + x2), we can take the derivative of both sides to get 1 2x(2x ln(2) + 2x) f (x) = 2 ln(2x + x2)+ , f(x) 2x + x2 so that   2x(2x ln(2) + 2x) f (x)= 2 ln(2x + x2)+ (2x + x2)2x. 2x + x2 Today we apply our ‘inverse ’ machinery to study the inverses of trigonometric functions. Before we begin, let’s remind ourselves quickly of some of the basics; if you feel that you need a more in-depth review, Appendix A of the book contains all of the material that you need to know.

Trigonometric Functions The basic trigonometric functions are defined in terms of coordinates of points on the unit . The point on the unit circle which is θ radians (in the counterclockwise direction) from (1, 0) has coordinates (cos(θ), sin(θ)).

θ sin(θ) We define tan( )= cos(θ) ; this can be thought of as the slope of the line joining the origin to the point (cos(θ), sin(θ)). You should make sure that you know all of the standard values for these basic trig functions, as well as their ‘co-’ versions 1 1 1 cos(θ) csc(θ)= , sec(θ)= , cot(θ)= = . sin(θ) cos(θ) tan(θ) sin(θ)

63 If you can familiarize yourself with the way in which these six trig functions are defined in terms of the unit circle, as well as the ‘standard’ triangles

then you can discern the many standard values for these trig functions; if you are uncom- fortable with this, then you should practice by working through problems in Appendix A. Inverse Trigonometric Functions As promised, we’re interested in inverting these trigonometric functions. However, a quick glance will indicate that there is no hope of inverting these functions because they are not one-to-one (i.e. their graphs don’t pass the horizontal line test). θ √1 Indeed if we try to solve some , say sin( )= 2 , then we get infinitely many θ ...,− 7π , − 5π , π , 3π , 9π , 11π ,... answers: = 4 4 4 4 4 4 . We should have expected this from the graph of y =sin(x), which intersects the line y √1 = 2 at infinitely many points:

y y √1 = 2

fx(x) = sin(x)

To get around this, we will artificially restrict the domain of these functions so that they become one-to-one. Practically speaking, this has the same effect as saying 1 “Find all values of θ in a certain interval such that sin(θ)=√ .” 2 If we choose our interval carefully, then we will always get a unique answer θ.

Example 75 (arcsin(x)). Let f(x) = sin(x). To make f(x) one-to-one, we restrict the f − π , π domain of to [ 2 2 ]. Other choices are possible, but this is the one that we use by convention. The important thing about this interval is that the range of f is still [−1, 1]. Then this function f is one-to-one, so it has an inverse function f −1(x) = arcsin(x). By construction, Domain of arcsin(x): [−1, 1],

64  π π  Range of arcsin(x): − , . 2 2 We sketch the graph of y = arcsin(x)andy = f(x) = sin(x) (with its restricted domain) below: y

f(x) = arcsin(x)

f(x) = sin(x)

x

To compute the derivative of arcsin(x), we use the general nonsense that we proved earlier about the derivatives of inverse functions to see that 1 1 (arcsin(x)) = = √ . cos(arcsin(x)) 1 − x2 This last equality follows from looking at the following triangle, which we construct to have an equal to arcsin(x):

√1 π Example 76. Returning to our earlier question, we see that arcsin( 2 )= 4 unambiguously. − − π This is very nice. Similarly, we can compute values like arcsin(0) = 0 and arcsin( 1) = 2 − π − by noting that sin(0) = 0 and sin 2 = 1.

Example 77 (arccos(x)). Just as with sin(x), we can define an inverse to cos(x)ifwe restrict its domain properly. In this case, we restrict the domain to [0,π]. As above, we get an inverse function arccos(x) to (this restricted version of) cos(x):

Domain of arccos(x): [−1, 1],

65 Range of arccos(x): [0,π]. We sketch the graph of y = arccos(x)andy =cos(x) (with its restricted domain) below:

f x x ( ) = arccos( ) y

x

f(x) = cos(x)

Using the same technique as before, we can show that −1 (arccos(x)) = √ . 1 − x2 Example 78 (arctan(x)). Finally we to tan(x). In this case, we restrict the domain − π , π x to ( 2 2 ). As above, we get an inverse function arctan( ) to (this restricted version of) tan(x):

Domain of arctan(x): (−∞, +∞),  π π  Range of arctan(x): − , . 2 2 We sketch the graph of y = arctan(x)andy =tan(x) (with its restricted domain) below:

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