Because x + y + z = 2π, then x = 2π - y - z
So cos(x) = cos(2π - y - z) = cos(2π - (y + z))
Difference formula for cosine is : cos(A-B) = cosA cosB + sinA sinB
So for our cos(2π - (y + z) ) = cos(2π) cos(y+z) + sin(2π) sin(y+z) = (1) (cos(y+z)) + (0) (sin(y+z))
So cos(x) = cos(y+z) and we can replace every cos(x) with cos(y+z) in the determinant.
Now the determinant looks like
| cos(y+z) cosz 1 |
| cosy 1 cosz |
| 1 cosy cos(y+z) |
"Solve" the determinant:
cos(y+z) [ cos(y+z) - cosycosz ] - cosz [ cosy cos(y+z) - cosz ] + (1) [ cosy cosy - 1 ]
cos2(y+z) - cos(y+z) cosy cosz - cosz cosy cos(y+z) + cos2z + cos2y - 1
The sum formula for cosine: cos(A+B) = cosA cosB - sinAsinB
Use this on each of the cos(y+z)'s
(cosycosz - sinysinz)2 - (cosycosz - sinysinz)cosycosz - coszcosy(cosycosz - sinysinz) + cos2z + cos2y - 1
"FOIL" the first term, and distribute all of the others:
cos2y cos2z - 2 sinysinzcosycosz + sin2y sin2z - cos2y cos2z + sinysinzcosycosz - cos2y cos2z + sinysinzcosycosz + cos2z + cos2y -1
Notice that there are NEGATIVE 2 "siny sinz cosy cosz" terms and two POSITIVE "siny sinz cosy cosz" terms, so they will cancel out.
Notice also that there both a positive and a negative "cos2y cos2z" terms, so they will disappear.
Now we have:
sin2y sin2z - cos2y cos2z + cos2z + cos2y -1
One of the trig identities is: sin2A + cos2A = 1, so it can be rearranged into sin2A = 1 - cos2A
Doing this to all of our sin2 terms, we get
(1-cos2y)(1-cos2z) - cos2y cos2z + cos2z + cos2y -1
FOIL the first two items:
1 - cos2z - cos2y + cos2y cos2z - cos2y cos2z + cos2z + cos2y - 1
Notice there is a positive and a negative cos2z term. Then cancel.
There is a positive and a negative cos2y term. They cancel.
And there is a positive cos2y cos2z as well as a negative one, so they cancel.
And what we are left with is 1 - 1 = 0
And we have just shown that the determinant =0 when x+y+z = 2π