Please help me with following:
cos^2 (pi/2 + x) + sin^2 (pi/2  x)
This is a bit different and I don't know how to get round it.
Help...
Thank you
Jorge

Jorge
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 14012005 20:47

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 14012005 20:53
(Original post by Jorge)
Please help me with following:
cos^2 (pi/2 + x) + sin^2 (pi/2  x)
This is a bit different and I don't know how to get round it.
Help...
Thank you
Jorge
Let Y = pi/2  x
cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1  cos2Y]/2
Can you pick it up from there?
Euclid. 
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 3
 14012005 20:59
(Original post by Jorge)
Please help me with following:
cos^2 (pi/2 + x) + sin^2 (pi/2  x)
This is a bit different and I don't know how to get round it.
Help...
Thank you
Jorge
sin(pi/2x) = cos x
and
cos(pi/2  (x)) = cos(x) = cos x
then it simplifies somewhat more quickly. 
Jorge
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 14012005 21:54
Thanks
I forgot to say that what the exercice wants is
to simplify
The answer is 1, and I'm afraid I can't get there,
Let X = pi/2 + x
Let Y = pi/2  x
cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1  cos2Y]/2
Can you pick it up from there?
Euclid.
1  sin^2 x + sin^2 y
sin^2 (xy) =1
But I don't think this is what they want...
Help...
Jorge 
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 5
 14012005 22:06
(Original post by Jorge)
Thanks
I forgot to say that what the exercice wants is
to simplify
The answer is 1, and I'm afraid I can't get there,
cos^2 x + sin^2 y
1  sin^2 x + sin^2 y
sin^2 (xy) =1
But I don't think this is what they want...
Help...
Jorge
Let Y = pi/2  x
cos^2(X) = [1 + cos2X]/2
sin^2(Y) = [1  cos2Y]/2
=> cos^2(X) + sin^2(X) = [1 + cos2X]/2 + [1  cos2Y]/2
=> [1 + cos2(pi/2 + x)]/2 + [1  cos2(pi/2  x)]/2
=> [1 + cos(pi + 2x)]/2 + [1  cos(pi  2x)]/2
=> [1 + cos(pi)cos2x  sin(pi)sin2x]/2 + [1  cos(pi)cos2x + sin(pi)sin2x]/2
you know that cos(pi) = 1, and sin(pi) = 0
=> [1  cos2x]/2 + [1 + cos2x]/2
=> sin^2(x) + cos^2(x)
=> 1
Euclid. 
Jorge
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 14012005 23:04
Euclid
Are you using the double angle formulae?
Because this exercice is prior to that, and I'm still a bit lost.
Jorge 
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 14012005 23:11
(Original post by Jorge)
Euclid
Are you using the double angle formulae?
Because this exercice is prior to that, and I'm still a bit lost.
Jorge
Euclid. 
Jorge
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 15012005 10:52
Hello Euclid
Thanks for your help
I'm still wondering as to the route the book wants you to follow.
At this stage  the addition formulae  I think that either I use it, or take advantage of cos^2A + sin^2A = 1
Now, as the result is simply: 1, I have a feeling that is what they're after.
Except that I am not too sure about the contents of the parentheses.
I suppose that as you are multiplying neg * pos you probably end up with nothing.
Could youy help me along these lines?
(This is a problem on page 57 of Collin's Pure 2, for those who might have the book)
Thanks
Jorge 
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 9
 15012005 11:59
(Original post by Jorge)
Hello Euclid
Thanks for your help
I'm still wondering as to the route the book wants you to follow.
At this stage  the addition formulae  I think that either I use it, or take advantage of cos^2A + sin^2A = 1
Now, as the result is simply: 1, I have a feeling that is what they're after.
Except that I am not too sure about the contents of the parentheses.
I suppose that as you are multiplying neg * pos you probably end up with nothing.
Could youy help me along these lines?
(This is a problem on page 57 of Collin's Pure 2, for those who might have the book)
Thanks
Jorge
Euclid. 
Jorge
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 15012005 12:02
I guess the addition fomulae and cos^2A + sin^2A = 1
From other questions in the same group it looks like they want me to know that things like cos pi/3 = 0.5, etc.. which makes life difficult for me.
Jorge 
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 11
 15012005 12:04
(Original post by Jorge)
I guess the addition fomulae and cos^2A + sin^2A = 1
From other questions in the same group it looks like they want me to know that things like cos pi/3 = 0.5, etc.. which makes life difficult for me.
Jorge
Euclid. 
Jorge
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 15012005 12:15
(Original post by Euclid)
I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names
Euclid.
Same problem here!
Addtion formulae:
sin(A+B) = sinAcosB + sinBcosA
rings a bell?
Jorge 
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 13
 15012005 12:21
(Original post by Jorge)
Same problem here!
Addtion formulae:
sin(A+B) = sinAcosB + sinBcosA
rings a bell?
Jorge
The using the addition formula:
cos(pi/2 + x) = cos[pi/2]cosx  sin[pi/2]sinx
sin(pi/2  x) = sin[pi/2]cos(x) + cos[pi/2]sin(x)
using the fact that cos90 = 0, and sin90 = 1, the equation reduces to:
cos(pi/2 + x) = cos[pi/2]cosx  sin[pi/2]sinx = sinx
sin(pi/2  x) = sin[pi/2]cos(x) + cos[pi/2]sin(x) = cos(x) = cosx
You are then left with:
(sinx)^2 + (cosx)^2
which you know is equal to one.
Euclid. 
Jorge
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 15012005 12:34
(Original post by Euclid)
Ahh right, they used to call it the double angle formulae when I was doing it.
You have: cos^2 (pi/2 + x) + sin^2 (pi/2  x)
The using the addition formula:
cos(pi/2 + x) = cos[pi/2]cosx  sin[pi/2]sinx
sin(pi/2  x) = sin[pi/2]cos(x) + cos[pi/2]sin(x)
using the fact that cos90 = 0, and sin90 = 1, the equation reduces to:
cos(pi/2 + x) = cos[pi/2]cosx  sin[pi/2]sinx = sinx
sin(pi/2  x) = sin[pi/2]cos(x) + cos[pi/2]sin(x) = cos(x) = cosx
You are then left with:
(sinx)^2 + (cosx)^2
which you know is equal to one.
Euclid.
What's confusing me here is the use of squares.
The additition formulae are for cos(A+B) and here is cos^(A+B)
Am I to understand that you proceed in the same fashion as if it were not squared and simply square the result at the end?
Thanks
Jorge 
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 15012005 12:35
(Original post by Jorge)
Thanks Euclid
What's confusing me here is the use of squares.
The additition formulae are for cos(A+B) and here is cos^(A+B)
Am I to understand that you proceed in the same fashion as if it were not squared and simply square the result at the end?
Thanks
Jorge
cos^2(A+B) = cos(A+B)*cos(A+B)
Do one of them and then square it.
Euclid. 
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 15012005 12:54
(Original post by Euclid)
Yes because
cos^2(A+B) = cos(A+B)*cos(A+B)
Do one of them and then square it.
Euclid.
Ahhhhh.....
Thanks
Jorge 
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 17
 15012005 13:00
((cos((Pi/2)+x))^2)+((sin((Pi/2)x))^2)
cos((Pi/2)+x)=cos(Pi/2)cosxsin(Pi/2)sinx=sinx
sin((Pi/2)x)=sin(Pi/2)cosxsinxcos(Pi/2)=cosx
((cos((Pi/2)+x))^2)+((sin((Pi/2)x))^2)=((sinx)^2)+((cosx)^2)=((sinx)^2)+( (cosx)^2)
=1 Q. E. D.
Newton.
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