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Prove sin(x) + cos(x) /=/ 1 Watch

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    (Original post by Asklepios)
    True but that's not really a triangle is it. But I agree stupid question unless you exclude 0 and 2pi from the domain.
    x = 2pi, x = 4pi, x = 6pi, x = 8pi, x = 10pi, x = 12pi, x =14pi, ... I could literally go on forever.

    The question does not specify anything about a range for x or it being in a triangle. Trigonometric functions are not by default restrained to triangles.
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    (Original post by Zacken)
    No, your question makes 0 sense.
    Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P

    (Original post by Zacken)
    Neither of you make any sense.

    1. sin x + cos x DOES EQUAL 1. You can't prove it doesn't because it does.
    Shoudln't both sin and cos be squared here?
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    (Original post by Mihael_Keehl)
    Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P



    Shoudln't both sin and cos be squared here?
    Ayman was wrong and he admitted it himself, read his edited post.

    sin x + cos x does equal 1 for certain values of x.
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    (Original post by Zacken)
    x = 2pi, x = 4pi, x = 6pi, x = 8pi, x = 10pi, x = 12pi, x =14pi, ... I could literally go on forever.

    The question does not specify anything about a range for x or it being in a triangle. Trigonometric functions are not by default restrained to triangles.
    I'll go back to the biology forums
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    (Original post by Mihael_Keehl)
    Another good way, thank you - I hadn't thought of doing this.

    Any reason why you chose 2sinxcosx?

    Any other value could be added right?
    It's just a bit of algebra to get it in the form a^2 + 2ab + b^2, but it solutions do exist for 2pi, and other periodic values.
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    (Original post by Asklepios)
    I'll go back to the biology forums
    In fact, you have sin x and cos x defined over the whole complex plane.
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    (Original post by Mihael_Keehl)
    Shoudln't both sin and cos be squared here?
    cosx+sinx=1 is true when x is a multiple of 2pi.

    It's a pointless question because k2pi is literally the answer to finding x.

    It's like saying x=7, prove that x is not not 7 or something.
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    (Original post by Kvothe the arcane)
    cosx+sinx=1 is true when x is a multiple of 2pi.

    It's a pointless question because k2pi is literally the answer to finding x.
    Not just that, x = 2pi k + pi/2 works as well.
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    (Original post by Mihael_Keehl)
    Ayman and asklepios answered what I was looking for, sorry if I did not construe the question well enough :P



    Shoudln't both sin and cos be squared here?
    Nah mate, I actually proved it otherwise. All periodic values yield a solution.
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    What if x is zero?
    Am I missing something?
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    (Original post by tazarooni89)
    What if x is zero?
    Am I missing something?
    Nopes, the OP and several others are.

    x can be any multiple of 2pi or x = 2pi k + pi/2 for any integer k.
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    (Original post by raman_17)
    Whats that about 3+4=5??
    I think she is referring to a 3, 4, 5 triangle perhaps.
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    (Original post by XxKingSniprxX)
    I think she is referring to a 3, 4, 5 triangle perhaps.
    cheers for that
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    (Original post by Mihael_Keehl)
    well for 3^2 + 4^2 = 5^2 square root everything.



    I see.

    So just by contradiction then?

    Nothing we can do by induction or anytihng else.
    nope........
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    Im late... Oh well
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    (Original post by Zacken)
    x = 2pi, x = 4pi, x = 6pi, x = 8pi, x = 10pi, x = 12pi, x =14pi, ... I could literally go on forever.

    The question does not specify anything about a range for x or it being in a triangle. Trigonometric functions are not by default restrained to triangles.
    ffs I wrote it misleadingly in the OP by accident. I was confused what you were saying which is probably in part due to the inarticulatirity of how I wrote the question.

    I understand what you were saying now, could you please refer to the edited OP.

    (Original post by aymanzayedmannan)
    It's just a bit of algebra to get it in the form a^2 + 2ab + b^2, but it solutions do exist for 2pi, and other periodic values.
    Ah yes, I see it know.

    (Original post by XxKingSniprxX)
    I think she is referring to a 3, 4, 5 triangle perhaps.
    Yes I was
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    (Original post by Mihael_Keehl)
    ffs I wrote it misleadingly in the OP by accident. I was confused what you were saying which is probably in part due to the inarticulatirity of how I wrote the question.

    I understand what you were saying now, could you please refer to the edited OP.



    Ah yes, I see it know.



    Yes I was
    I'll just refer to my initial "proof" to show you why this is not the case. The range of sin2x falls between -1 and 1. This implies that sinx + cosx can take up any real value between -sqrt(2) and sqrt(2) so a solution of 1 must exist.
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    (Original post by aymanzayedmannan)
    I'll just refer to my initial "proof" to show you why this is not the case. The range of sin2x falls between -1 and 1. This implies that sinx + cosx can take up any real value between -sqrt(2) and sqrt(2) so a solution of 1 must exist.
    Thanks yes, I see
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    Ok, given you're excluding certain domains, you may write sinx+cosx as (root2)sin(x+pi/4), which means sin(x+pi/4)=1/root2, the inverse of such will give answers pi/4 or 3pi/4, which means x=pi/2 or 0, given I presume these values are excluded in the domain then there is no solution for sinx+cosx=1

    I can't be bothered to write it more rigorously but that's what you need.
 
 
 
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