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Re: [obm-l] Trigonometria




sen^2 x + sen^2 y = 1 - cospi/5
sen^2y -(1-sen^2x)=-cospi/5
(seny-cosx)(seny+cosx)=-cospi/5
 
(cos(90-y)-cosx)(cos(90-y)+cosx)=-cospi/5
 
cos(x+y) =cosxcosy-senxseny
cos(x-y)=cosxcosy+senxseny
 
cos(x+y)+cos(x-y)=2cosxcosy
cos(x-y)-cos(x+y)=2senxseny
 
x+y=a
x-y=b
x=(a+b)/2
y=(a-b)/2
cosa+cosb=2cos(a+b)/2cos(a-b)/2
cosb-cosa=2sen(a+b)/2sen(a-b)/2
 
(cos(90-y)-cosx)(cos(90-y)+cosx)=-cospi/5
2sen(90+x-y)/2 *sen(90-(x+y))/2 * 2*cos(90+x-y)/2*cos(90-(x+y))/2=-cospi/5
2sen(90+x-y)/2 * cos(90+x-y)/2 * 2sen(90-(x+y))/2cos(90-(x+y))/2=-cospi/5
sen(90 +x-y)sen(90-(x+y))=-cospi/5
cos(x-y)cos(x+y)=-cospi/5
x + y = pi/5
cos(x-y)=-1
x-y=npi  (n=,,,-3,-1,-,3,5,7..)
x + y = pi/5
2x= pi*(n+1/5)
x=pi*(n/2 +1/10)
2y=pi*(1/5 - n)
y = pi(1/10 -n/2)
 
x=pi*(n/2 +1/10)
y = pi(1/10 -n/2)
 (n=,,,-3,-1,-,3,5,7..)
 
abraço, saulo.
 
 
On 1/17/06, Klaus Ferraz <klausferraz@yahoo.com.br> wrote:
Determine os valores de x e y que satisfazem as equacoes:
 
x + y = pi/5
sen^2 x + sen^2 y = 1 - cospi/5


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