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   Author  Topic: Trig Equation  (Read 712 times)
ThudnBlunder
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Trig Equation  
« on: Oct 21st, 2008, 2:30pm »
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Find a necessary and sufficient condition on n so that the equation 2sin+ ncos = 7
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Grimbal
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Re: Trig Equation  
« Reply #1 on: Oct 21st, 2008, 3:20pm »
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... has a solution for some ?
 
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Michael Dagg
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Re: Trig Equation  
« Reply #2 on: Oct 21st, 2008, 4:12pm »
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As Grimbal noted, the result can't hold for all  
alpha .  I smell a good algebraic
discussion, let me guess: roots of unity...
« Last Edit: Oct 21st, 2008, 4:14pm by Michael Dagg » IP Logged

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Michael Dagg
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Re: Trig Equation  
« Reply #3 on: Oct 21st, 2008, 4:26pm »
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That is, if n is an integer.  Otherwise it is trivial.
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ThudnBlunder
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Re: Trig Equation  
« Reply #4 on: Oct 21st, 2008, 5:50pm »
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on Oct 21st, 2008, 4:26pm, Michael Dagg wrote:
That is, if n is an integer.  Otherwise it is trivial.

n and yes, is restricted somewhat.
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teekyman
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Re: Trig Equation  
« Reply #5 on: Oct 21st, 2008, 11:48pm »
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n needs to be at least 3?
« Last Edit: Oct 22nd, 2008, 11:31am by teekyman » IP Logged
ThudnBlunder
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Re: Trig Equation  
« Reply #6 on: Oct 22nd, 2008, 12:36pm »
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on Oct 21st, 2008, 11:48pm, 1337b4k4 wrote:
n needs to be at least 3?

Yes. was that a guess?
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Re: Trig Equation  
« Reply #7 on: Oct 22nd, 2008, 12:45pm »
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nope.
a*sin() + b*cos() = (a+ b) * sin(+ ) for = tan(b / a). Thus n would have to be at least 3 since sin(+ ) is at most 1. Also, |n| needs to be at least 3 would be more correct.
« Last Edit: Oct 22nd, 2008, 12:50pm by teekyman » IP Logged
Michael Dagg
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Re: Trig Equation  
« Reply #8 on: Oct 23rd, 2008, 1:09pm »
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One doesn't see that identity all that often.
 
Also, with a little calculus you can show that the  
range of the function  f(x) = (sqrt 7 - 2 sin x)/cos x  
is the complement of the open interval  (-sqrt 3,sqrt 3)  .
« Last Edit: Oct 23rd, 2008, 1:13pm by Michael Dagg » IP Logged

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