<p> Ch 6 Inverse Circular Functions and Trigonometric Equations</p><p>6.1 Inverse Circular Functions</p><p>Functions</p><p> a x x a</p><p> b y y b</p><p> c c</p><p>A function is a one-to-one function if, for elements c and d in the domain of f, c  d implies f c  f d </p><p>(different inputs always result in different outputs) Graphical</p><p>If 2,5 lies on the graph of f, then 5,2 lies on the graph of f 1.</p><p> f : a  b f 1 :b  a</p><p>a,b on the graph of f  b,a on the graph of f 1 The Horizontal Line Test</p><p>A function is one-to-one if every horizontal line intersects the graph of the function at most once.</p><p>Inverse function</p><p>The inverse function of a one-to-one function f is defined as follows: f 1  y, xx, y f </p><p>1 Notation: f 1  f</p><p>Finding an inverse function 1) Verify f is a one-to-one function 2) Solve y  f x for x in terms of y. 3) Interchange x and y to obtain y  f 1x. 4) Check your work. f  f 1x x? f 1 f x  x? We often restrict the domain of a function that is not one-to-one to make it one-to-one without changing the range.</p><p>To find f 1x from f x, follow these steps:</p><p>1 Replace f x with y and interchange x and y. 2 Solve for y. 3 Replace y with f 1x.</p><p>Inverse circular functions are not one-to-one.</p><p>Inverse sine function y  sin 1 x or y  arcsin x means that   x  sin y for   y  2 2 1 Find y  arcsin 2 Find y  sin 11</p><p>Find y  sin 1 2 Inverse cosine function y  cos1 x or y  arccos x means that x  cos y for 0  y  </p><p>Find y  arccos1</p><p> 2  Find y  cos1   2  Inverse tangent function y  tan 1 x or y  arctan x means that   x  tan y for   y  2 2</p><p>Find   arctan1 in degrees.</p><p>Inverse Cotangent, Secant, and Cosecant y  cot 1 x or y=arccot x means that x  cot y for 0  y   y  sec1 x or y=arcsec x means that  x  sec y for 0  y   , y  2 y  csc1 x or y  arccsc x means that   x  csc y for   y  , y  0 2 2 Find   sec1 2 in degrees.</p><p>Calculator: 1 sec1 x can be evaluated as cos1 x 1 csc1 x can be evaluated as sin 1 x cot 1 x can be evaluated as 1 tan 1 x  0  x  1 180  tan1 , x  0  x Find y  csc1 3 in radians.</p><p>Find   cot 1 0.3541 in degrees.</p><p> 1 3 Find sin tan .  2  1 5  Find tancos  .   13</p><p> 1 Find cosarctan 3  arcsin .  3  2 Find tan2arcsin .  5</p><p>Find sintan 1 u.</p><p>Find cos2sin 1 u. The optimal angle of elevation θ that a shot- putter should aim for has been modeled as:  v2    arcsin   2   2v  64h  where v is the velocity of the shot and h is the initial height of the shot. Suppose a shot-putter is consistent with h  6.6ft and v  42ft/sec. 6.2 Trigonometric Equations I</p><p>Solve 2sin 1  0 over0,360; all solutions.</p><p>Solve sin tan  sin over0,360 Solve tan 2 x  tan x  2  0 over0,2 </p><p>Find all solutions of cot xcot x  3 1 Solve tan x  3  sec x over0,2  Solving a Trigonometric Equation 1. Decide whether the equation is linear or quadratic in form, so you can determine the solution method. 2. If only one trigonometric function is present, first solve the equation for that function. 3. If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve. 4. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval. 5. Try using identities to change the form of the equation. It may be helpful to square each side of the equation first. In this case, check for extraneous solutions. 6.3 Trigonometric Equations II</p><p> x Solve 2sin 1 over0,2 ; all solutions. 2</p><p>Solve cos2x  cos x over0,2 . Solve 4sin cos  3 over0,360; all solutions.</p><p>Solve tan 3x  sec3x  2 over0,2 . The “A” key above middle C on a piano has </p><p> the fundamental frequency f1  440Hz with </p><p> associated pressure P1  0.002sin880t. The </p><p> string also vibrates at f2  880Hz, f3 1320 Hz, etc. Corresponding pressures are 0.002 0.002 P  sin1760t, P  sin 2640t , 2 4 3 3 etc. The graph of P  P1  P2  P3  P4  P5 is saw-toothed.</p><p>Find max value of P and where this value occurs?</p><p>A graphing calculator shows that the maximum value of P is approximately 0.00317. The maximum occurs at t = x ≈ 0.000191, 0.00246, 0.00474, 0.00701, and 0.00928. 6.4 Equations Involving Inverse Trigonometric Functions</p><p>   Solve y  3cos 2x, x  0, .  2</p><p>Solve 2arcsin x   .</p><p>1 Solve cos1 x  sin 1 . 2  Solve arcsin x  arccos x  . 6</p>