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    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
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    (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 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.
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    (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
    If you use

    sin(pi/2-x) = cos x

    and

    cos(pi/2 - (-x)) = cos(-x) = cos x

    then it simplifies somewhat more quickly.
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    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.
    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
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    (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 X = pi/2 + x
    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.
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    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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    (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
    In that case it is probably best to use the fact that cos(pi/2 + x) = -sinx and sin(pi/2 - x) = cosx.

    Euclid.
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    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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    (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
    OK., what identities are you allowed to use at this stage for answering that question? Then we'll worth with them.

    Euclid.
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    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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    (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
    I'm very sorry but remind me what the addition fomulae formuale are. I never remember the names

    Euclid.
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    (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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    (Original post by Jorge)
    Same problem here!

    Addtion formulae:

    sin(A+B) = sinAcosB + sinBcosA

    rings a bell?

    Jorge
    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.
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    (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.
    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
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    (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
    Yes because

    cos^2(A+B) = cos(A+B)*cos(A+B)

    Do one of them and then square it.

    Euclid.
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    (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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    ((cos((Pi/2)+x))^2)+((sin((Pi/2)-x))^2)

    cos((Pi/2)+x)=cos(Pi/2)cosx-sin(Pi/2)sinx=-sinx
    sin((Pi/2)-x)=sin(Pi/2)cosx-sinxcos(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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