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March 12th, 2008, 07:52 AM  #1 
Newbie Joined: Mar 2008 Posts: 6 Thanks: 0  range of the sin function
find the domain and range of y=Sqrt(1sinx) why is the range: (0<=y<=Sqrt2) why does the range go to Sqrt2, why not 1? 
March 12th, 2008, 08:31 AM  #2 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Greetings: Domain: The sqrt function is defined for all positive reals. Because the sine function never exceeds a value of 1, it follows that 1sin(x) >= 0 for all real numbers x. Hence sqrt(1sin(x)) exists for all real x. Restated, the domain is 'all real numbers'. Range: Since sin(x) has a maximum value of 1, 1sin(x) >= 0. Thus sqrt(1sin(x)) >= 0. Since sin(x) has a munimum of 1, 1sin(x) can never exceed 1(1)=2. Thus sqrt(1sin(x)) <= sqrt(2). Conclusion: 0 <= sqrt(1sin(x)) and sqrt(1sin(x)) <= sqrt(2) implies a range between 0 and sqrt(2) inclusive. Regards, Rich B. rmath4u2@aol.com[/color] 
March 12th, 2008, 12:04 PM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
The extreme points of sin(x) are 1 and 1, so 1sin(x) is in [0, 2]. The square root maps this to [0, sqrt(2)].

March 13th, 2008, 08:20 PM  #4 
Newbie Joined: Mar 2008 Posts: 6 Thanks: 0 
thanks for the answers.. long live infinity 

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function, range, sin 
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