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March 12th, 2008, 07:52 AM   #1
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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?
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March 12th, 2008, 08:31 AM   #2
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[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]
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March 12th, 2008, 12:04 PM   #3
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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)].
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March 13th, 2008, 08:20 PM   #4
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thanks for the answers..

long live infinity
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