MHB Can you prove the cosine rule for three angles in a triangle?

[anemone]

For all $x,\,y,\,z \in R$ with $x+y+z=2\pi$, prove that $\cos^2 x+\cos^2 y+\cos^2 z+2\cos x\cos y \cos z=1$

 

[kaliprasad]

For all $x,\,y,\,z \in R$ with $x+y+z=2\pi$, prove that $\cos^2 x+\cos^2 y+\cos^2 z+2\cos x\cos y \cos z=1$

Spoiler

$\cos²x+\cos²y+\cos²z+2 \cos x \cos y \cos z $
= $cos²x+cos²y+cos²(\pi-z)+2 cos x cos y cos z$
= $cos²x+cos²y+cos²(x+y)+2 cos x cos y cos z$
= $cos²x+cos²y+(cos x cos y - sin x sin y)^2+2 cos x cos y cos z$
=$ cos²x+cos²y+ cos^2 x cos^2 y + sin ^2 x sin^2 y- 2 cos 2 x cos y sin x sin y+2 cos x cos y cos z$
= $cos²x+cos²y+ cos^2 x cos^2 y + ( 1- cos ^2 x)(1- cos ^2 y)- 2 cos 2 x cos y sin x sin y+2 cos x cos y cos z$
= $cos ^2 x + cos^2 y + cos^2 x cos^2 y + 1 – cos^2 x – cos^2 y + cos^2 x cos^2 y - 2 cos 2 x cos y sin x sin y+2 cos x cos y cos z$
= $1+ 2 cos^2 x cos^2 y - 2 cos 2 x cos y sin x sin y+2 cos x cos y cos z$
= $1 + 2 cos x cos y ( cos x cos y – sin x sin y)+2 cos x cos y cos z$
= $1 + 2 cos x cos y cos (x+y)+2 cos x cos y cos z$
= $1 – 2 cos x cos y cos (\pi – ( x + y))+2 cos x cos y cos z$
= $1 – 2 cos x cos y cos z+2 cos x cos y cos z$
= 1

 

[anemone]

Thanks for participating, kaliprasad!

Spoiler

Another method would be to perceive the given equation as a quadratic equation $k^2+(2\cos y \cos z)k +(\cos^2 y+\cos^2 z-1)=0$ and our task is to show that this quadratic equation has a root $k=\cos x$.

By the quadratic formula, we have

$\begin{align*}k&=\dfrac{-2\cos y \cos z\pm\sqrt{4\cos^2 y \cos^2 z-4(\cos^2 y+\cos^2 z-1)}}{2}\\&=-\cos y \cos z\pm\sqrt{(1-\cos^2y)(1-\cos^2z)}\\&=-\cos y \cos z\pm|\sin y\sin z|\end{align*}$

Since $-\cos y \cos z+\sin y\sin z=-\cos(y+z)=\cos(\pi-x)=\cos x$, we find that $k=\cos x$ satisfies the quadratic equation, as desired.

 

[kaliprasad]

Spoiler

$\cos^2x+\cos^2y+\cos^2z+2\cos x\cos y\cos z=\cos^2x+\cos^2y+\cos^2(\pi−z)+2\cos x\cos y\cos z$
$=\cos^2x+\cos^2y+\cos^2(x+y)+2\cos x\cos y\cos z$
$=\cos^2x+\cos^2y+(\cos x\cos y−\sin x\sin y)^2+2\cos x\cos y\cos z$
$=\cos^2x+\cos^2y+\cos^2x\cos^2y+\sin^2x\sin^2y−2\cos x \cos y\sin x\sin y+2\cos x\cos y\cos z$
$=\cos^2x+\cos^2y+\cos^2x\cos^2y+(1−\cos^2x)(1−\cos^2y)−2\cos x\cos y\sin x\sin y+2\cos x\cos y\cos z$
$=\cos^2x+\cos^2y+\cos^2x\cos^2y+1–\cos^2x–\cos^2y+\cos^2x\cos^2y−2\cos x\cos y\ sin x\sin y+2\cos x\cos y\cos z$
$=1+2\cos ^2x \cos^2y−2\cos^2x\cos y\sin x\sin y+2\cos x\cos y \cos z$
$=1+2\cos x\cos y(\cos x\cos y–\sin x\sin y)+2\cos x\cos y\cos z$
$=1+2\cos x\cos y\cos(x+y)+2\cos x\cos y\cos z$
$=1–2\cos x\cos y\cos(π–(x+y))+2\cos x\cos y\cos z$
$=1–2\cos x\cos y\cos z+2\cos x\cos y\cos z=1$

rewrote the solution with the latex as previous solution to not clear to read