My Math Forum range of the sin function

 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top