ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Zebra on April 21, 2011, 09:08:46 pm
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We are told that the domain of Cos(x) is 0 to pie
Cos(sin-1(x))
the implied domain?
Going through questions I got wrong in Exercise 3C is driving me mad!
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since the range of sin^-1(x) is -pi/2 to pi/2, for the function Cos(sin^-1(x)) to exist, we must restrict the domain of sin^-1(x) to x=0 to x=1. Hence the implied domain is 0≤x≤1
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So, we need find the implied domain of
])
as you said,  = [0,\pi])
You may remember from composite functions in methods that 
= 
However we must restrict this range so that it becomes a subset of the domain of cos(x), so we get
0 ≤
≤ 
EDIT: Solve for x
sin(0) ≤ x ≤ sin(
)
which is
0 ≤ x ≤ 1
Is this working out right? :S
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So, we need find the implied domain of ])
as you said,  = [0,\pi])
You may remember from composite functions in methods that 
= 
However we must restrict this range so that it becomes a subset of the domain of cos(x), so we get
0 ≤
≤ 
Multiply everything by
so middle becomes x
sin(0) ≤ x ≤ sin(
)
which is
0 ≤ x ≤ 1
Is this working out right? :S
Yes - ur working is correct. Except, your not multiplying everything by sin....
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Yes - ur working is correct. Except, you're (:P) not multiplying everything by sin....
Oh my bad, was I meant to say I was solving for x? Because I think that's what I was meant to do...
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Yes - ur working is correct. Except, you're (:P) not multiplying everything by sin....
Oh my bad, was I meant to say I was solving for x? Because I think that's what I was meant to do...
LOL! - Haha... I actually laughed at how you edited my quote. Good one!
P.S. I would love to know you in real life :P :P :P