Inverse Trig Functions

Transcription

1 MATH 7 Inverse Trig Functions Dr. Neal, WKU A function y = f (x) is one to one if it is always the case that different x values are assigned to different y values. For example, y = x + 4 is one to one, but y = sin x is not one to one. The sine function assigns many different x values to the same y. For instance, π, π, π, etc. all have a sine of. y = x + 4 is one-to-one y = sin x is not one-to-one A one to one function f (x) always has an inverse denoted by f (x). The trigonometric functions cos x, sin x, tan x are not one-to-one; thus technically they do not have inverses. However we can make them one-to-one by restricting the domains to only two quadrants so that we have only one segment of the graph. The Inverse Cosine (or Arccos) With cos x, we restrict the domain to [, π] (the st and nd quadrants). Then the inverse cosine exists: For x, cos x is the angle whose cosine is x, where the angle must be in either the st or nd quadrant and must be written in radians. f (x) = cos x Dom: x π Range: y f (x) = cos x Dom: x Range: y π 5 4 We know the cosine values of the standard angles around the unit circle: 4 cos() = cos = cos = 4 cos = cos = cos cos 4 = cos 5 = = cos() =

2 So we can compute the inverse cosine, or arccos, of the values,,,,,,,, : cos x is the angle between and π whose cosine is x cos () = cos = cos = 4 cos = cos () = cos = cos = 4 cos = 5 cos () = Evaluating cos(cos x) and cos (cos x) It is always the case that cos(cos x ) = x for x. This is because cos x is the angle whose cosine is x ; thus, cos(cos x ) = x. We dont even have to calculate cos x in this case. For example, cos( cos ( / ) ) = and cos( cos ( / ) ) = ( ) and cos( cos (4)) are undefined because 5 and 4 are not within Also, cos cos (5) the interval x. On the other hand, cos (cos x ) = x only for x. In this case, we can start with any angle x. Then cos x. So cos (cos x ) exists, but cos (cos x ) must be an angle from to π and may not equal the original angle x.

3 Example. Compute (a) cos cos 7 ( (b) cos cos 5 ( (c) cos cos 4 ((. Solution. We first evaluate the cosine of the inner given angle, then evaluate the inverse cosine of the result. The answers must all be angles from to π. (a) cos cos 7 ( = cos ( = 5 (b) cos cos 5 ( = cos ( = (c) cos cos 4 (( = cos ( = 4 The Inverse Sine (or Arcsin) With sin x, we restrict the domain to [ π/, π/] (the st and 4th quadrants). Now the inverse sine exists: For x, sin x is the angle whose sine is x, where the angle must be in either the st or 4th quadrant. If the angle is in the 4th quadrant, then we write it as a negative angle. f (x) = sin x Dom: x Range: y f (x) = sin x Dom: x Range: y 4 We know the sine values of the standard angles from π/ to π/: sin ( = sin ( = sin ( = 4 sin ( = ( ) = sin sin = 4 sin = 4 sin = sin =

4 So now we can compute the inverse sine, or arcsin, of the values,,,,,,,, : Dr. Neal, WKU sin x is the angle between and whose sine is x sin () = sin = sin = 4 sin = sin ( ) = sin = sin = 4 sin = sin ( ) = Evaluating sin(sin x) and sin (sinx) It is always the case that sin(sin x ) = x for x. This is because sin x is the angle whose sine is x ; thus, sin(sin x ) = x. We dont even have to calculate sin x in this case. For example, sin( sin ( / ) ) = and sin( sin ( / ) ) = ( ) and sin( sin (. 4) ) are undefined because and.4 are not within Also, sin sin () the interval x. On the other hand, sin (sin x ) = x only for x. In this case, we can start with any angle x. Then sin x. So sin (sin x ) exists, but sin (sin x ) must be an angle from π/ to π/ and may not equal the original angle x.

5 Example. Compute (a) sin sin 4 ( (b) sin sin 5 4 ( (c) sin sin 7 ( (. Solution. We first evaluate the sine of the inner given angle, then evaluate the inverse sine of the result. The answers must all be angles from π/ to π/. (a) sin sin 4 ( = sin ( = (b) sin sin 5 ( 4 = sin ( = 4 (c) sin sin 7 ( ( = sin ( = The Inverse Tangent (or Arctan) With tan x, we restrict the domain to ( π/, π/) (the st and 4th quadrants). Then the inverse tangent exists: For all x, tan x is the angle whose tangent is x, where the angle must be in either the st or 4th quadrant. If the angle is in the 4th quadrant, then we write it as a negative angle. f (x) = tan x Domain: < x < Range: < y < f (x) = tan x Domain: < x < Range: < y <

6 4 We know the tangent values of the standard angles from π/ to π/: tan ( = tan ( = tan 4 ( = tan ( = ( ) = tan tan = 4 tan 4 = tan = tan = So now we can compute the inverse tangent, or arctan, of the values,,,,,,,, : tan x is the angle between and whose tangent is x tan () = tan ( ) = tan () = 4 tan = tan ( ) = tan = tan ( ) = 4 tan ( ) = tan () = Evaluating tan(tan x) and tan (tan x) It is always the case that tan (tan x ) = x for all x. This is because tan x is the angle whose tangent is x ; thus, tan (tan x ) = x. We dont even have to calculate tan x in this case. For example, tan ( tan (.4)) =. 4 and tan ( tan (. 5) ) =. 5

7 On the other hand, tan (tan x ) = x only for < x <. In this case, we can start with any angle x. Then < tan x <. So tan (tan x ) exists, but tan (tan x ) must be an angle between π/ and π/ and may not equal the original angle x. Example. Compute (a) tan tan 5 ( (b) tan tan 7 ( (c) 4 tan tan ((. Solution. We first evaluate the tangent of the inner given angle, then evaluate the inverse tangent of the result. The answers must all be angles from π/ to π/. (a) tan tan 5 ( = tan ( = (b) tan tan 7 ( = tan 4 () = 4 (c) tan tan (( = tan ( ) =