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 Algebra Pre-Algebra and Basic Algebra Math Forum

 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(1-sinx) 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 1-sin(x) >= 0 for all real numbers x. Hence sqrt(1-sin(x)) exists for all real x. Restated, the domain is 'all real numbers'. Range: Since sin(x) has a maximum value of 1, 1-sin(x) >= 0. Thus sqrt(1-sin(x)) >= 0. Since sin(x) has a munimum of -1, 1-sin(x) can never exceed 1-(-1)=2. Thus sqrt(1-sin(x)) <= sqrt(2). Conclusion: 0 <= sqrt(1-sin(x)) and sqrt(1-sin(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 1-sin(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 Tags function, range, sin Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post th3j35t3r Number Theory 5 November 17th, 2013 01:47 PM happy21 Calculus 9 August 17th, 2013 07:31 AM sachinrajsharma Calculus 2 April 8th, 2013 10:09 PM balbasur Algebra 1 May 30th, 2010 04:28 PM Tartarus Algebra 8 November 24th, 2009 01:50 PM

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