## anonymous one year ago Cos^2(x)+cos^2(y)+cos^2(z)=1 prove the triangle is right

• This Question is Open
1. Loser66

*

2. anonymous

For a triangle $$x+y+z=\pi$$ and we get$\cos^2 x+\cos^2 y+\cos^2 z=1 \\ \cos^2 x+\cos^2 y+\cos^2 (\pi-x-y)=1 \\ \cos^2 x+\cos^2 y+\cos^2 (x+y)=1 \\ \cos^2 x+\cos^2 y=1-\cos^2 (x+y) \\ \cos^2 x+\cos^2 y=\sin^2 (x+y)=(\sin x \cos y+\cos x \sin y)^2 \\ \cos^2 x+\cos^2 y=\sin^2 x \cos^2 y+\cos^2 x \sin^2 y + 2 \sin x \sin y \cos x \cos y \\ \cos^2 x (1-\sin^2 y)+\cos^2 y(1-\sin^2 x)=2 \sin x \sin y \cos x \cos y \\ \cos^2 x \cos^2 y= \sin x \sin y \cos x \cos y \\ \cos x \cos y(\cos x \cos y-\sin x \sin y)=0 \\ \cos x \cos y \cos (x+y)=0 \\ \cos x \cos y \cos (\pi-x-y)=0 \\ \cos x \cos y \cos z=0$so one of the angles must be equal to $$\frac{\pi}{2}$$

3. anonymous

you were not here, and I gave the answer directly, sry

4. anonymous

@mukushla no problem thx for the answer but man yesterday after posting it i continued thinking of it and after itching my mind alott yessss i knew it I was really happy 😂😂 look thats my prove

5. anonymous

Here

6. anonymous

Sorry i cut a part cause it didnt fit

7. anonymous

Quite right! Note that in the last part you cannot divide by $$2 \cos x \cos y$$, see last part of my solution.