{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 55 "High School Modul es > Trigonometry by Gregory A. Moore " }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 256 40 "Simplifying Inverse Function Expressions" }} {PARA 0 "" 0 "" {TEXT -1 126 "\nSimplifying various expressions involv ing inverse sine, cosine, and tangents - and a visualization of the \" mysterious case.\"\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : E xecute the Code Resource section first. Although there will be no outp ut immediately, these definitions are used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 8 " 0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; with(pl ots): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1151 "ArcsinTracePlot := proc(t)\n local a,c1,c2,c3,A,B,C,E,F,G,GG,H,J;\n theta||1 := t; \n theta||2 := arcsin( sin(t));\n a := sin(theta||1):\n \+ c1 := COLOR(RGB, 1, .75, .2 ):\n c2 := COLOR(RGB, .7, .4, .1):\n \+ c3 := COLOR(RGB, 1, .2, .1):\n A := plot(sin(x), x = -Pi..2*P i, color = c3, tickmarks = [1,1]):\n B := plot(sin(x), x = -Pi/2..Pi/2, color = c2, thickness = 4):\n C := polygonplot( \+ [[-Pi/2,-1],[Pi/2,-1],[Pi/2,1],[-Pi/2,1],[-Pi/2,-1]], \n \+ color = c1, style = patchnogrid ):\n E := plot( [[-Pi/2,-1],[Pi/ 2,-1],[Pi/2,1],[-Pi/2,1],[-Pi/2,-1]], \n color = orange, linestyle = 3):\n F := plottools[arrow]( [theta||1, a], [theta|| 2,a], \n .05, .10, .15, color = red):\n G := plott ools[arrow]( [theta||1, 0], [theta||1,a], \n .15,.25,.10 , color = red):\n GG := plottools[arrow]( [theta||2,a], [theta||2 , 0], \n .15,.25,.10, color = red): \n H := textpl ot([theta||1, -.15, evalf(theta||1,3)],align=\{ABOVE, CENTER\}):\n \+ J := textplot([ -.2, a, evalf(a, 3)],align=\{ABOVE, LEFT\}):\n d isplay([F,G,GG, A,B,C,E,H,J ]);\nend proc :" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 36 " 1. Simple Inverse Trig Expressions" }}{PARA 0 "" 0 " " {TEXT -1 93 "\nA simple inverse trig expression involves find an ang le which gives a specific value.\n\n " }{TEXT 264 11 "Example 1.1 " }{TEXT -1 11 " : Compute " }{XPPEDIT 18 0 "arcsin( sqrt(3)/2 )" "6#- %'arcsinG6#*&-%%sqrtG6#\"\"$\"\"\"\"\"#!\"\"" }{TEXT -1 61 " \n\nIn th is case, we are trying to find an angle theta, where " }{XPPEDIT 18 0 "sin(theta) = sqrt(3)/2" "6#/-%$sinG6#%&thetaG*&-%%sqrtG6#\"\"$\"\"\" \"\"#!\"\"" }{TEXT -1 25 ". Perhaps you can guess \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "sin(theta) = sqrt(3)/2;\nsolve(%, theta); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "arcsin( sqrt(3)/2);" }} }{PARA 0 "" 0 "" {TEXT -1 60 "\nIn other cases, it's not simple to gue ss the answer.\n\n " }{TEXT 267 11 "Example 1.2" }{TEXT -1 11 " : \+ Compute " }{XPPEDIT 18 0 "arcsin( 2/3 )" "6#-%'arcsinG6#*&\"\"#\"\"\" \"\"$!\"\"" }{TEXT -1 2 " \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "arcsin( 2/3); evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 7 "\n\n \+ " }{TEXT 268 12 " Example 1.3" }{TEXT -1 239 " : Simplify arcsin( 2 ) \n\nNotice that the only valid input to arcsine and arccosine are num bers in the range [-1,1]. Maple will solve equations for values outsid e that range using more advanced mathematics (a branch of mathematics \+ called " }{TEXT 270 16 "Complex Analysis" }{TEXT -1 88 ", which you wi ll study if you major in mathematics) and comes up with a complex numb er.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "arcsin( 2); evalf( %);" }}}{PARA 0 "" 0 "" {TEXT -1 8 " \n\n " }{TEXT 266 11 "Example 1.4" }{TEXT -1 13 " : Simplify " }{XPPEDIT 18 0 "arccos( 2/3 ) " "6# -%'arccosG6#*&\"\"#\"\"\"\"\"$!\"\"" }{TEXT -1 1 "\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "arccos( 2/3); evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 6 "\n " }{TEXT 265 11 "Example 1.5" }{TEXT -1 28 " : S implify arctan( 100 ) \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "arctan( 100 ); evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 47 " 2. Compose Function with Its Inverse \+ Function" }}{PARA 0 "" 0 "" {TEXT -1 304 "\nThis is the simpler of two cases. Inverse trig functions are always one-to-one and yield unique \+ angles. When a trig function is applied to that angle, there is a uniq ue result. In other words, a trig function composed with its inverse c ancel one another's transformations, and render the original value." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "That is, sin(arcsin(x)) = x, in the same way that " }{XPPEDIT 18 0 "exp(ln(x)) = x" "6#/-%$expG6#-%#lnG6#%\"xGF*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "(sqrt(x))^2 = x" "6#/*$-%%sqrtG6#%\"xG\"\"#F(" }}{PARA 0 "" 0 "" {TEXT -1 6 "\n " }{TEXT 262 7 "Example" }{TEXT -1 33 " : Simplify \+ sin( arcsin( .819 ))\n" }}{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sin( arcsin( .819 ));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "And while we're at it..." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "cos( arccos( .819 ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "tan( arctan( .819 ));" }}}{PARA 0 " " 0 "" {TEXT -1 1 " " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 51 " 3. Com pose Function with Another Inverse Function" }}{PARA 0 "" 0 "" {TEXT -1 71 "\nWhat about a trig function applied to another inverse trig fu nction?\n\n" }{TEXT 261 19 " Example 3.1 " }{TEXT -1 25 " : sin( arccos( .819) )\n\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "sin( arccos( .819 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 108 " \nMaple gives us the answer promptly. However, to better understand the process, we ca n follow these steps.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "c := cos( arccos( .819 ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Well_Known_Formula := sin(x)^2 + cos(x)^2 = 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "subs( cos(x) = c, Well_Known_Formula);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "soln := solve( %, sin(x) ); " }}}{PARA 0 "" 0 "" {TEXT -1 283 "\nWe have a value for sin(x) ... in fact, two values... which is one answer too many. How do we tell whic h of the two possible answers is the real answer?\n\nA positive cosine = .819 is in the first quadrant, rather than the second quadrant. The sine in the first quadrant is positive.\n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 32 "arccos( .819 ) < Pi/2; evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "abs( soln[1]);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "\n\nThis type of problem can be solved even when dealing with u nknowns. \n\n" }{TEXT 263 18 " Example 3.2" }{TEXT -1 26 " : ta n( arcsin( x/3 ) )\n\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " t an( arcsin( x/3 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 155 "\nTo solve this problem on paper, its best to draw a triangle which has an opposite s ide of length x and a hypotenuse of 3. Such an angle will have sine : \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "sin(theta) = x/3;" }}} {PARA 0 "" 0 "" {TEXT -1 155 "\nThen we find the third side of the tri angle using the theorem of Pythagoras. Then the tangent is easy to fin d as the opposite side over the adjacent side." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "third_side := sqrt( 3^2 - x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "tan(theta) = x/third_side;" }}} {PARA 0 "" 0 "" {TEXT -1 31 "\n\n\n Here are other examples ..." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "tan( arcsin( x/3 ) );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "cos( arctan( x/3 ) );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sin( arccos( x/3 ) );" }}}} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 43 " 4. Compose Inverse Function wit h Function" }}{PARA 0 "" 0 "" {TEXT -1 171 "\nThis is the more mysteri ous case. It seems to work well in some cases - where the arcsine undo es the work of the sin. But in other cases it does not seem to work as well." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "arcsin( sin( Pi/6) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "arcsin( sin( 7*Pi/6) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "arcsin( sin( 19*Pi/6 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 205 "\n\n\nThe reason for this is t hat the sine can accept any real number as its input. If we were to us e the sine with restricted domain, the inverse would cancel the functi on. The restricted domain for sine is " }{XPPEDIT 18 0 "[-Pi/2, Pi/2] " "6#7$,$*&%#PiG\"\"\"\"\"#!\"\"F)*&F&F'F(F)" }{TEXT -1 43 ". Any valu e in that domain will work well.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "arcsin( sin( Pi/3) );\narcsin( sin( -Pi/5 ) );\narcsi n( sin( 0 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 37 "\n\nBut any angle outs ide this domain, " }{XPPEDIT 18 0 "[-Pi/2, Pi/2]" "6#7$,$*&%#PiG\"\"\" \"\"#!\"\"F)*&F&F'F(F)" }{TEXT -1 41 " will not come back to where it \+ started.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "arcsin( sin( 2 *Pi/3) );\narcsin( sin( -5*Pi/6 ) );\narcsin( sin( 2*Pi ) );" }}} {PARA 0 "" 0 "" {TEXT -1 76 "\n\nHere is a diagram which demonstrates \+ why arcsin( sin( 7Pi./6)) = - Pi/6.\n\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ArcsinTracePlot(7*Pi/6);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "ArcsinTracePlot(5*Pi/6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ArcsinTracePlot(-5*Pi/6);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 50 " 5. More Compl icated Inverse Function Expressions" }}{PARA 0 "" 0 "" {TEXT -1 50 "\n We can also do some more complicated problems. \n\n" }{TEXT 260 24 " \+ Example 5.1 " }{TEXT -1 47 " : Simplify cos( arcsin( 3/4) + \+ arcsin( 2/3) )\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "cos( arc sin( 3/4) + arcsin( 2/3) );" }}}{EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "theta := arcsin(3/4); \ntheta := evalf(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "phi := arcsin(2/3);\nphi := \+ evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "cos( theta + p hi);" }}}{PARA 0 "" 0 "" {TEXT -1 189 "\nWhile that gives us the answe r, there is another way of doing this problem which might work out bet ter for paper-and-pencil solving - plus it gives us an exact radical f orm of the answer. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "Cosi neAdditionIdentity := \n cos( alpha + beta) = cos(alpha)*cos(b eta) - sin(alpha)*sin(beta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "subs( \{ alpha = arcsin( 3/4), beta = arcsin( 2/3) \}, CosineAd ditionIdentity );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "Now we have reduced the one hard problem to four simple problems that we have alr eady seen how to solve." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 " simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(rhs(% ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT 259 40 " \n \251 20 02 Waterloo Maple Inc" }}}{MARK "0 1" 23 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }