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Math Help - Problem 20

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    Problem 20

    Proof that a triangle with acute angles x,y is right if and only if:
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Proof that a triangle with acute angles x,y is right if and only if:
    ok, so i think i have one of the implications down, but i'm having trouble with the converse. tell me if what i did is right so far.

    Proof:
    Assume x and y are the acute angles of a right triangle, then x + y = 90. This means that x and y are compliments. Recall that the sine of an angle is equal to the cosine of its compliment, and thus we can let cosy = sinx and cosx = siny.

    Now sin(x + y) = sinxcosy + sinycosx = sinx(sinx) + siny(siny) = sin^2x + sin^2y

    Now for the converse, we use the contrapositive. Assume sin(x + y) not= sin^2x + sin^2y....?
    Last edited by Jhevon; March 5th 2007 at 06:20 PM.
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    Quote Originally Posted by Jhevon View Post
    Now for the converse, we use the contrapositive. Assume sin(x + y) not= sin^2x + sin^2y....?
    No.

    If sin(x+y)=sin^2x+sin^2y then the triangle is right, given x,y are the two acute angles.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    No.

    If sin(x+y)=sin^2x+sin^2y then the triangle is right, given x,y are the two acute angles.
    isn't that what we are supposed to prove? i dont think you can just state it like that
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    Wouldn't that mean proving that sin^2(x)+sin^2(y)=1
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    I stole this problem and solution of the internet.

    The only if part is easy to show.

    If a triangle is right then:
    sin^2 x + sin^2 y = sin^2 x + cos^2 x =1
    And,
    sin(x+y)=sin(90)=1.
    Thus,
    sin^2 x + sin^2 y = sin(x+y)

    Q.E.D.

    ---
    Now the forward condition.

    Assume, (where x,y are acute),
    sin^2 x + sin^2 y = sin(x+y)
    Then,
    sin^2 x + sin^2 y = sin x*cos y+cos x*sin y

    Move stuff around,
    sin x(sin x - cos y) = sin y(cos x - sin y)

    Now, we are going to solve this in a very strange way.
    Note that both sin x and sin y are positive.

    Thus, sgn (sin x - cos y) = sgn (cos x - sin y)

    Where sgn is the "sign function sgn(x)", we define it to be +1 for x>0, 0 for x=0, and -1 for x<0.*

    Meaning, since sin x and sin y are both positive the second factors have the same sign (or zero).

    Case 1: sgn (sin x - cos y) = +1
    In that case sin x - cos y>0.
    But then sgn (cos x - sin y) = +1 that is cos x - sin y>0.
    We therefore have,
    sin x > cos y and cos x < sin y.
    Square both sides (they are non-negative),
    sin^2 x > cos^2 y and cos^2 x> sin^2 y.
    Add,
    1=sin^2 x+cos^2 x > cos^2 y+sin^2 y=1.
    A contradiction!

    Case 2: sgn (sin x - cos y)=-1
    In this case we arrive at 1<1 and have the same situation if we followed the reasoning in Case 1.

    Case 3: sgn(sin x - cos y)=0
    This is what remains.
    This must be true if the premesis is true.
    Thus,
    sin x = cos y.
    Thus,
    x+y=90
    Because x,y are acute by conditions of the problem.

    *)NOTE: This is not the "sign function" as in the derivative of y=|x|. I just made it up to make this proof smoother.
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