Proof that a triangle with acute angles x,y is right if and only if:

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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

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....?

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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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran

isn't that what we are supposed to prove? i dont think you can just state it like that

Wouldn't that mean proving that sin^2(x)+sin^2(y)=1

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Quick's Prime Property:
For
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and the ratio of numbers which is indivisible by any whole number
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but
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is
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then
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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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"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran