inversetrig - The inverse sine func-on: sin ­1(x)...

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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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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.

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