Unformatted text preview: The inverse sine funcon: sin
1(x) also called arcsin(x) Queson The funcon y=sin(x) is inverble. True False Answer False The funcon y=sin(x) is not one
to
one, so it’s not inverble So… What do we mean by “inverse sine” funcon? In order to deﬁne the inverse of sine, we must ﬁrst restrict the funcon to an interval where it’s one
to
one. What’s a good choice for this interval?
3π/2
π/2 0 π/2 3π/2 Is [0,2π] a good choice? Namely, is y=sin(x) inverble on [0,2π]?
3π/2
π/2 0 π/2 3π/2 Is [
π/2, π/2] a good choice? YES!!! y=sin(x) inverble on [
π/2, π/2]. YES: y=sin(x) inverble on [
π/2, π/2].
3π/2
π/2 0 π/2 3π/2 Take inverse y=sin(x) y=sin
1(x) y=sin
1(x) Take inverse Connuous Diﬀerenable Domain: [
π/2, π/2] Range: [
1,1] Connuous Diﬀerenable Domain:[
1,1] Range: [
π/2, π/2] π/2 y=sin
1(x) y=x y=sin(x)
π/2 π/2
π/2 y=sin
1(x) Deﬁned for
1≤ x ≤ 1 Takes values between –π/2 and +π/2 y=sin
1(x) Deﬁned for
1≤ x ≤ 1 Takes values between –π/2 and +π/2 If y=sin
1(x), then sin(y) = x sin
1(x) is the only angle between –π/2 and +π/2 whose sine is = x sin
1(x) is the only angle between –π/2 and +π/2 whose sine is = x • sin
1(1/2) = π/6 • sin
1(1) = π/2 • sin
1(√2/2) = π/4 • sin
1(√3/2) = π/3 …… sin
1(x) is the only angle between –π/2 and +π/2 whose sine is = x What’s sin
1(
1)? A point on the unit circle has coordinates (cosθ, sinθ). To ﬁnd sin
1(
1) we must look for a point with y
coordinate
1 This suggests θ=(3/2)π but it’s not a good choice, because sin
1 must take values between –π/2 and +π/2. Start from (3/2)π, and add ±2π, unl you get into the desired range. sin
1 (
1)=
π/2 Find sin
1(
√3/2). Start from (5/3)π, and add ±2π, unl you ﬁnd an angle between –π/2 and +π/2 sin
1 (
√3/2)=
π/3 Problem: Compute cos(sin
1(1/3)) and tan(sin
1(1/3)) …without a calculator! Note: there is no “famous” angle with sine = 1/3, so we can’t guess. Review of trig Problem: Compute cos(sin
1(1/3)) and tan(sin
1(1/3)) =1 =1/3 θ
sin(θ)= a/c = (1/3) / 1 = 1/3 Phytagorean theorem cos(θ)=b/c = b = (c2
a2)1/2 = (1
1/9) ½=(8/9) ½ =(2√2)/3 tan(θ)=a/b = (1/3)/(2√2/3) = 1/(2√2) Problem: Compute cos(sin
1(1/3)) and tan(sin
1(1/3)) =1 we could have worked =1/3 directly with the (similar) triangle: θ
=3 =1 θ ‘ The inverse sine funcon: cos
1(x) also called arccos(x) y=cos(x) not one
to
one We resct the domain to [0,π] to make it inverble y=cos(x) y=cos(x) y=sin
1(x) y=cos
1(x) 0 π/2 π Take inverse Connuous Diﬀerenable Domain: [0, π] Range: [
1,1] Connuous Diﬀerenable Domain:[
1,1] Range: [0, π] y=cos
1(x) Deﬁned for
1≤ x ≤ 1 π Takes values between 0 and π π/2 1 1 π/2 π cos
1(x) is the only angle between 0 and π whose cosine is = x • cos
1(1/2) = π/3 • cos
1(1) = 0 • cos
1(√2/2) = π/4 • cos
1(√3/2) = π/6 • cos
1(0) = π/2 • cos
1(
1/2) = 2π/3 • cos
1(
√2/2) = 3π/4 • cos
1(
√3/2) = 5π/6 The inverse tangent funcon: tan
1(x) also called arctan(x) y=tan(x) not one
to
one We resct the domain to [
π/2,π/2] to make it inverble y=tan(x) y=cos(x) y=sin
1(x) y=tan
1(x) 0 π/2
π/2 π π/2 Take inverse Connuous Diﬀerenable Domain: (
π/2, π/2) Range: (
∞,+∞) Connuous Diﬀerenable Domain:(
∞,+∞) Range: (
π/2, π/2) ...
View
Full
Document
This note was uploaded on 12/07/2010 for the course MATH MATH 2B taught by Professor Famiglietti,c during the Fall '09 term at UC Irvine.
 Fall '09
 FAMIGLIETTI,C
 Calculus

Click to edit the document details