Show that the determinant of | Row1: cosx cosz 1; Row 2: cosy 1 cosz; Row 3: 1 cosy cosx | = 0 when x + y + z = 2 pi

Question

Show that the determinant of

| cosx cosz 1 |

| cosy 1 cosz |

| 1 cosy cosx |

is = 0 when x + y + z = 2 pi

Answer 1

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:

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