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

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    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect

    EDIT - I made a mess of this question. What I meant to say is that:

    sin^2(x) + cos^2(x) = 1 for all values of x

    However

    sin(x) + cos(x) /=/ 1 for all values of x.

    Prove why this is not the case.
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    (Original post by Mihael_Keehl)
    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect
    you are not making sense...
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    (Original post by Mihael_Keehl)
    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect
    Whats that about 3+4=5??
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    (Original post by Mihael_Keehl)
    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect
    Is your question: prove that \sin x + \cos x \neq 1?

    If so, x = 2 \pi n or \displaystyle x = 2\pi n + \frac{\pi}{2} for any n \in \mathbb{Z} is an easy counterexample.
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    (Original post by TeeEm)
    you are not making sense...
    (Original post by raman_17)
    Whats that about 3+4=5??
    well for 3^2 + 4^2 = 5^2 square root everything.

    (Original post by Zacken)
    Is your question: prove that sin x + cos x \neq 1?

    If so, x = 2 \pi n or \displaystyle x = 2\pi n + \frac{\pi}{2} for any n \in \mathbb{Z} is an easy counterexample.
    I see.

    So just by contradiction then?

    Nothing we can do by induction or anytihng else.
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    Just sub in a random value for x which doesn't make either one 0 or 1.
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    (Original post by raman_17)
    Whats that about 3+4=5??
    3^2+4^2=5^2 \nRightarrow 3+4=5
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    just substitute a random number in and prove by counter example.
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    (Original post by Mihael_Keehl)
    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect
    Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

    So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1 (unless X=0).

    Sin(X) + cos(X) = (O + A)/H.

    If you think of the hypotenuse as the shortest distance between two points, A+O must be more than H.
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    (Original post by Vikingninja)
    Just sub in a random value for x which doesn't make either one 0 or 1.
    (Original post by Mihael_Keehl)
    well for 3^2 + 4^2 = 5^2 square root everything.
    Neither of you make any sense.

    1. sin x + cos x DOES EQUAL 1. You can't prove it doesn't because it does.
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    I am still lost ...

    Name:  Z5.jpg
Views: 139
Size:  10.2 KB


    I am sure Zacken will help here
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    (Original post by cubic^3)
    just substitute a random number in and prove by counter example.
    (Original post by Asklepios)
    Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

    So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1.
    x = 2\pi...
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    (Original post by Mihael_Keehl)
    I get the whole 3+4 /=/ 5 argument

    but from sin^2(x) + cos^2(x) = 1

    how do we establish a proof to show that it is incorrcect
    Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

    This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

    This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.

    Edit: nah, I was wrong.
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    (Original post by aymanzayedmannan)
    Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

    This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

    This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.
    x = 2\pi.
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    (Original post by TeeEm)
    I am still lost ...

    Name:  Z5.jpg
Views: 139
Size:  10.2 KB


    I am sure Zacken will help here
    I've got no clue what's going on here either.
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    (Original post by Asklepios)
    Remember sin(X) = opposite/hypotenuse and cos(X) = adjacent/hypotenuse

    So you can see why the sum of squares is 1 by Pythagoras' theorem, but for the sum itself it can never be 1.

    Sin(X) + cos(X) = (O + A)/H.

    If you think of the hypotenuse as the shortest distance between to points, A+O must be more than H.
    Perfect. Thank you.

    (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.
    lol?

    if x=30 then that yields 0.5(1 + sqrt3)

    Did you mean sin^2x + cos^2 x
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    (Original post by Mihael_Keehl)
    Perfect. Thank you.



    lol?

    if x=30 then that yields 0.5(1 + sqrt3)

    Did you mean sin^2x + cos^2 x
    No, your question makes 0 sense.
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    (Original post by aymanzayedmannan)
    Add 2sinxcosx to both sides, you get sin^2(x) + 2sinxcosx + cos^2(x) = 1 + 2sinxcosx

    This can be rewritten as (sinx + cosx)^2 = 1 + 2sinxcosx

    This implies that sinx + cosx = ±sqrt(1 + sin2x), a solution does not exist for any value of x. Maybe you could even graph the function.
    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?
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    (Original post by Zacken)
    x = 2\pi.
    True but that's not really a triangle is it. But I agree stupid question unless you exclude 0 and 2pi from the domain.
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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?
    No it's the non squared part of (sinx+cosx)^2
 
 
 
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