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

 

[CellCoree]

This is a well-known trigonometric identity known as the "cosine rule." It states that for any three angles in a triangle, the sum of the squares of the cosines of those angles plus twice the product of the cosines of the angles is equal to 1. This can be proven using the law of cosines and some basic algebraic manipulations. It is a fundamental and useful identity in trigonometry and has many applications in solving problems involving triangles and circles. This challenge serves as a good exercise in utilizing trigonometric identities and applying them in mathematical proofs.