Victoria V. answered • 09/09/19

15+ Years Experience Teaching / Tutoring Trigonometry

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 ]

Distribute:

cos^{2}(y+z) - cos(y+z) cosy cosz - cosz cosy cos(y+z) + cos^{2}z + cos^{2}y - 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) + cos^{2}z + cos^{2}y - 1

"FOIL" the first term, and distribute all of the others:

cos^{2}y cos^{2}z - 2 sinysinzcosycosz + sin^{2}y sin^{2}z - cos^{2}y cos^{2}z + sinysinzcosycosz - cos^{2}y cos^{2}z + sinysinzcosycosz + cos^{2}z + cos^{2}y -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 "cos^{2}y cos^{2}z" terms, so they will disappear.

Now we have:

sin^{2}y sin^{2}z - cos^{2}y cos^{2}z + cos^{2}z + cos^{2}y -1

One of the trig identities is: sin^{2}A + cos^{2}A = 1, so it can be rearranged into sin^{2}A = 1 - cos^{2}A

Doing this to all of our sin^{2} terms, we get

(1-cos^{2}y)(1-cos^{2}z) - cos^{2}y cos^{2}z + cos^{2}z + cos^{2}y -1

FOIL the first two items:

1 - cos^{2}z - cos^{2}y + cos^{2}y cos^{2}z - cos^{2}y cos^{2}z + cos^{2}z^{ } + cos^{2}y - 1

Notice there is a positive and a negative cos^{2}z term. Then cancel.

There is a positive and a negative cos^{2}y term. They cancel.

And there is a positive cos^{2}y cos^{2}z 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