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| Don't post to this addres |
 Posted: Thu Dec 18, 2003 1:08 pm |
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Guest
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Hi all,
How do you prove that the minimum value of
sin a cos b - cos a sin b
never goes below -1
Yours faithfully,
Euan |
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| Tomas Andersson |
| Posted: Thu Dec 18, 2003 1:14 pm |
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Guest
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Don't post to this address wrote:
Quote: Hi all,
How do you prove that the minimum value of
sin a cos b - cos a sin b
never goes below -1
Yours faithfully,
Euan
sin a cos b - cos a sin b = sin(a - b)
sin never goes below -1 |
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| Maxi |
| Posted: Thu Dec 18, 2003 1:38 pm |
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Guest
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sin a cos b - cos a sin b = sin(a-b)
:-)
--
Maxi |
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| Don't post to this addres |
| Posted: Fri Dec 19, 2003 9:02 am |
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Guest
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ok.good.Thank you guys.
how do you prove sine never goes below -1 ? |
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| Toni Lassila |
| Posted: Fri Dec 19, 2003 9:40 am |
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Guest
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On 19 Dec 2003 06:02:26 -0800, newsgroupeuan@hotmail.com (Don't post
to this address) wrote:
Quote: how do you prove sine never goes below -1 ?
By the geometric definition of sin as the ratio of the sides of a
right triangle. The triangle inequality says that this ratio can never
exceed 1.
Note however that it is possible to extend the definition of sin to
complex numbers in which case it's no longer bound. |
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| A N Niel |
| Posted: Fri Dec 19, 2003 10:49 am |
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Guest
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In article <4ea1f4c0.0312190602.16851d40@posting.google.com>, Don't
post to this address <newsgroupeuan@hotmail.com> wrote:
Quote: ok.good.Thank you guys.
how do you prove sine never goes below -1 ?
This depends on your definition of sine. What is yours?
With some definitions it is trivial, with other definitions
it requires some work.
For example, say we take this definition:
the solution of the D.E. y'' + y = 0, y(0)=0, y'(0)=1.
In this case you can use the trick: consider
u = y^2+(y')^2, show u'=0, so u is constant.
Therefore y^2 <=1. |
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