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The Inverse Trigonometric Functions Let's again review a few things about inverse functions. • To have an inverse function, a function must be one-to-one (remember if a horizontal line intersections the graph of a function in more than one place it is NOT one-to-one). •If we have points on a function graph and we trade x and y places we'll have points on the inverse function graph. •Functions and their inverses "undo" each other so f f 1 x •Since x and y trade places, the domain of the function is the range of the inverse and the range of the function is the domain of the inverse •The graph of a function and its inverse are reflections about the line y = x (a 45° line). Is y = sin x a one-to-one function? No! A horizontal line intersects its graph many times. If we only look at part of the sine graph where the x values go from -/2 to /2 and the y values go from -1 to 1, we could find an inverse function. If we want to find an inverse sine function, we can't have the sine repeat itself so we are only going to look at part of the sine graph. We are going to define the inverse function of sine then to be y = sin-1x. Remember for inverse functions x and y trade places so let's take the values we got for sin x and we'll trade them places for y = sin-1x with our restricted domain. x y = sin x x y = sin-1 x Remember we are only looking at x values from 1 1 2 2 -/2 to /2 so that we can 1 1 have a one-to6 2 2 6 one function 0 0 0 0 that will have an inverse. 1 1 -1 2 2 6 So for y = sin x, we can put in 6 numbers between -1 and 1. What we get out is the angle 1 1 2 2 between -/2 to /2 that has that sine value. Let's graph both of these. Here is the graph of sin x between -/2 to /2 x y = sin x 1 1 Notice they are 2 reflections 1 about a 45° line 6 2 0 6 2 0 1 2 1 2 1 Here is the graph of from -1 to 1 y = sin-1 x x 2 1 2 6 1 sin-1 2 x 0 0 1 2 6 1 2 Since sin x and sin-1 x are inverse functions they "undo" each other. f 1 f x sin sin x x where 1 x 2 2 1 1 f f x sin sin x x where 1 x 1 sin sin 6 6 1 CAUTION!!! You must be careful that the angle is between -/2 to /2 to cancel these. Looking at the unit circle, if we choose from -/2 to /2 it would be the right half of the circle. It looks like the When you see 1 3 1 3 , 0,1 , 2 2 answersine should 2 2 inverse then,be it 2 2 2 2 2 11/6 but , means they'll give 2 , 2 2 2 2 3 3 3 remember the 3 1 you the sine value 4 5 , 3 1 2 2 , 4 range what you 2 2 6 and it's or asking get out must 6 which angle onbe thean angle between -/2 right half of the unit 0 1,0 1,0 to /2, sothis we'd use circle has sine 2 a coterminal angle value. 7 in this range which 6 11 3 1 5 , 3 1 is: , 4 7 6 2 2 2 2 4 3 2 2 , 2 2 1 3 , 2 2 3 2 0,1 5 3 4 1 3 , 2 2 2 2 , 2 2 1 sin 6 2 1 Notice y values (sine repeat themselves This isthe asking, where on values) the rightnever half of the unit circle is in this half the circle so the sine function would be 1-to-1 here. the sine value -1/2? Remember when you have sin-1 x it means "What angle between -/2 to /2 has a sine value of x?" Let's look at the unit sin 1 2 2 4 circle to answer this. 1 3 , 2 2 0,1 2 3 3 5 4 6 2 2 2 , 2 3 1 , 2 2 1 3 , 2 2 2 3 2 2 2 , 2 3 1 2 ,2 4 6 3 sin 3 2 1 This is 5/3 but need 1,0 7 6 3 1 2 , 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 2 , 2 2 2 2 , 2 This has a a sine This is is asking, asking, "What "What angle angle from from -/2 -/2 to to /2 /2 has sine value 2 over 2?" value of of square negativeroot square root 3 over 2?" 3 sin 1 2 sin sin 2 4 4 1 Now This remember looks like sine for inverse and its inverse sign we should only use cancel values out on butthe then right half you'd of be thegetting unit circle. an So answer this isout asking, that was "Where not in the onrange the right so half you of must the be circle careful is the sine on these. value square Let's work root 2the over stuff 2?"in the parenthesis first and see what happens. 1 3 , 2 2 0,1 2 3 3 5 4 6 2 2 2 , 2 3 1 , 2 2 1 3 , 2 2 2 3 2 2 2 , 2 3 1 2 ,2 4 6 1,0 7 6 3 1 2 , 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 2 , 2 2 2 2 , 2 Let's think about an inverse cosine function. If we chose the right half of the unit circle for cosine values would we have a one-to-one function? NO! For example: 1 cos cos 3 3 2 Let's look at the graph of y = cos x and see where it IS one-to-one 1 3 , 2 2 0,1 2 3 3 5 4 6 2 2 2 , 2 3 1 , 2 2 1 3 , 2 2 2 3 2 2 2 , 2 3 1 2 ,2 4 6 1,0 7 6 3 1 2 , 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 From 0 to 5 3 11 7 6 4 1 3 , 2 2 3 1 2 , 2 2 2 2 , 2 So cos-1 x is the inverse function of cos x but the domain is between -1 to 1 and the range is from 0 to . f x cos cos x x where 0 x 1 1 f f x coscos x x where - 1 x 1 f 1 1 1 cos 2 3 3 5 1 cos 2 6 1 So this is asking where on the upper half of the unit circle does the cosine value equal 1/2. 1 3 , 2 2 0,1 2 3 3 5 4 6 2 2 2 , 2 3 1 , 2 2 1 3 , 2 2 2 3 2 2 2 , 2 3 1 2 ,2 4 6 1,0 7 6 3 1 2 , 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 2 , 2 2 2 2 , 2 tan-1 x is the inverse function of tan x but again we must have restrictions to have tan x a one-to-one function. f 1 f x tan tan x x where 1 x 2 2 1 1 f f x tan tan x x where x 1 3 , 2 2 0,1 2 3 3 5 4 6 2 2 2 , 2 3 1 , 2 2 1 3 , 2 2 2 3 2 2 2 , 2 3 1 2 ,2 4 6 1,0 7 6 3 1 2 , 2 We'll take tan x from -/2 to /2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 2 , 2 2 2 2 , 2 Definition: The Inverse Sine Function For 1 x 1 y sin 1 x provided sin y x and / 2 y / 2. Another notation for inverse sine is arcsin( x) which is read "arc sine of x" Definition: The Inverse Cosine Function For 1 x 1 y cos 1 x provided cos y x and 0 y . Another notation for inverse cosine is arccos( x) which is read "arc cosine of x" Definition: The Inverse Tangent Function For any real number x, y tan 1 x provided tan y x and / 2 y / 2. Another notation for inverse tangent is arctan( x) which is read "arc tangent of x" We've talked about inverse functions for sine, cosine and tangent and saw that we must restrict the range so we have a one-to-one function. Here is a summary: 1 sin x Restricted to angles on right half of unit circle listed from -/2 to /2 1 cos x Restricted to angles on upper half of unit circle listed from 0 to 1 tan x Restricted to angles on right half of unit circle listed from -/2 to /2 What about the inverses of the reciprocal trig functions? 1 csc x 1 sin x Restricted to angles on right half of unit circle listed from -/2 to /2 1 sec x 1 cos x Restricted to angles on upper half of unit circle listed from 0 to 1 cot x 1 tan x Restricted to angles on upper right half of unit half of circle unit circle listed from -/2 listed fromto0/2 to Cosecant and secant have same restrictions as their reciprocal functions but cotangent does not. Cosecant excludes 0 since it would be undefined there and secant excludes /2. We've seen in this section how to work some of these if the values are exact values on the unit circle and if not how to use the calculator. Now we need to find exact values even if not on the unit circle. This means the 1 1 o 2 2 tan cos 2 2 3 a 1 cosine of some angle (with reference angle ) is -1/3 so label adjacent and hypotenuse. 3 Inverse cosine is quadrants I 2 2 and II and since the cosine value is negative, the -1 terminal side must be in drop down line to x axis quadrant II. The secret is to draw a triangle in the correct quadrant and label the sides you know, figure out the other side by Pythagorean Theorem, and then find the trig function you need. If you need the exact value of a reciprocal function you can find the "flip" over the value you have and find the inverse of the reciprocal function (being careful that you meet the restrictions). 1 2 sec 2 cos 3 2 1 1 1 3 , 2 2 0,1 2 2 3 3 5 4 6 2 2 2 , 2 This is asking what angle (on the upper half of the unit circle) has a cosine value of -1/2 or in other words, cos = -1/2 so what is ? 3 1 , 2 2 1,0 1 3 , 2 2 3 2 2 , 2 2 3 1 , 2 2 4 6 7 6 3 1 , 2 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 , 2 2 2 2 2 , 2 If you use your calculator to find one of the reciprocal functions you use the same strategy of "flipping" over the given value and then you can use the inverse trig function button for the reciprocal. 1 csc 5 sin 0.20 5 1 1 Now use your calculator and compute this. We've talked about inverse functions for sine, cosine and tangent and saw that we must restrict the range so we have a one-to-one function. Here is a summary: sin 1 Restricted to angles on right half of unit circle listed from -/2 to /2 cos 1 tan Restricted to angles on upper half of unit circle listed from 0 to Restricted to angles on right half of unit circle listed from -/2 to /2 1 What about the inverses of the reciprocal trig functions? csc sec cot sin 1 cos tan Restricted to angles on upper half of unit circle listed from 0 to Restricted to angles on upper right half of unit half of circle unit circle listed from -/2 listed fromto0/2 to 1 1 Restricted to angles on right half of unit circle listed from -/2 to /2 1 1 1 Cosecant and secant have same restrictions as their reciprocal functions but cotangent does not. For help on using your calculator to compute inverse trig functions, click here. Also in the video clips I show you how to use the calculator. We've seen in the last section how to work some of these if the values are exact values on the unit circle and if not how to use the calculator. Now we need to find exact values even if not on the unit circle. This means the 1 1 o 2 2 tan cos 2 2 3 a 1 x cosine of some angle x (with reference angle x’ ) is -1/3 so label adjacent and hypotenuse. 3 Inverse cosine is quadrants 2 2 x’ x I and II and since the cosine value is negative, the -1 terminal side must be in drop down line to x axis quadrant II. The secret is to draw a triangle in the correct quadrant and label the sides you know, figure out the other side by Pythagorean Theorem, and then find the trig function you need. If you need the exact value of a reciprocal function you can find the "flip" over the value you have and find the inverse of the reciprocal function (being careful that you meet the restrictions). 1 2 sec 2 cos 3 2 1 1 1 3 , 2 2 0,1 2 2 3 3 5 4 6 2 2 2 , 2 This is asking what angle (on the upper half of the unit circle) has a cosine value of -1/2 or in other words, cos = -1/2 so what is x? 3 1 , 2 2 1,0 1 3 , 2 2 3 2 2 , 2 2 3 1 , 2 2 4 6 7 6 3 1 , 2 2 0 1,0 2 5 4 4 3 2 2 2 , 2 1 3 , 2 2 3 2 0,1 5 3 11 7 6 4 1 3 , 2 2 3 1 , 2 2 2 2 2 , 2 If you use your calculator to find one of the reciprocal functions you use the same strategy of "flipping" over the given value and then you can use the inverse trig function button for the reciprocal. 1 csc 5 sin 0.20 5 1 1 Now use your calculator and compute this. Definition: The Inverse Cosecant, Secant and Cotangent Functions 1 csc x sin for x 1 x 1 -1 1 sec x cos for x 1 x 1 cot 1 -1 x 2 tan 1 x More on Cotangent -1 1 tan x -1 1 1 cot x tan x 2 for x 0 for x 0 for x 0 Function Gallery: Inverse Trigonometric Functions