# Math Help - Problem 20

1. ## Problem 20

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

2. Originally Posted by ThePerfectHacker
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....?

3. Originally Posted by Jhevon
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.

4. Originally Posted by ThePerfectHacker
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

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

6. 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.
1=sin^2 x+cos^2 x > cos^2 y+sin^2 y=1.

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.