**Unformatted text preview: **The inverse sine func-on: sin
1(x) also called arcsin(x) Ques-on The func-on y=sin(x) is inver-ble. True False Answer False The func-on y=sin(x) is not one
to
one, so it’s not inver-ble So… What do we mean by “inverse sine” func-on? In order to deﬁne the inverse of sine, we must ﬁrst restrict the func-on 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) inver-ble on [0,2π]?
3π/2
π/2 0 π/2 3π/2 Is [
π/2, π/2] a good choice? YES!!! y=sin(x) inver-ble on [
π/2, π/2]. YES: y=sin(x) inver-ble 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 Con-nuous Diﬀeren-able Domain: [
π/2, π/2] Range: [
1,1] Con-nuous Diﬀeren-able 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, un-l you get into the desired range. sin
1 (
1)=
π/2 Find sin
1(
√3/2). Start from (5/3)π, and add ±2π, un-l 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 func-on: cos
1(x) also called arccos(x) y=cos(x) not one
to
one We res-ct the domain to [0,π] to make it inver-ble y=cos(x) y=cos(x) y=sin
1(x) y=cos
1(x) 0 π/2 π Take inverse Con-nuous Diﬀeren-able Domain: [0, π] Range: [
1,1] Con-nuous Diﬀeren-able 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 func-on: tan
1(x) also called arctan(x) y=tan(x) not one
to
one We res-ct the domain to [
π/2,π/2] to make it inver-ble y=tan(x) y=cos(x) y=sin
1(x) y=tan
1(x) 0 π/2
π/2 π π/2 Take inverse Con-nuous Diﬀeren-able Domain: (
π/2, π/2) Range: (
∞,+∞) Con-nuous Diﬀeren-able Domain:(
∞,+∞) Range: (
π/2, π/2) ...

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- Fall '09
- FAMIGLIETTI,C
- Calculus, the00, π/200