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

Results 1 to 6 of 6

- March 5th 2007, 05:20 PM #1

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- March 5th 2007, 06:04 PM #2
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....?

- March 5th 2007, 07:07 PM #3

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- March 6th 2007, 09:46 AM #4

- March 6th 2007, 12:20 PM #5

- March 13th 2007, 09:35 AM #6

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

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.