How to Prove the Determinant of a Cosine Matrix?

[Nobody1111]

Proof, that determinant (with n rows and columns)

| cosx 1 0 0 ... 0 0 |
| 1 2cosx 1 0 ... 0 0 |
| 0 1 2cosx 1 ... 0 0 |
| 0 0 1 2cosx ... 0 0 | = cos nx
|...... |
| 0 0 0 0 ... 2cosx 1 |
| 0 0 0 0 ... 1 2cosx |

The main proble for me is to evaluate this determinant.

Do you have any ideas?

Thanks for help.

 

[NateTG]

It looks a lot like you want to use induction on the number of rows&columns.

 

[M98Ranger]

The determinant of a matrix can be calculated by multiplying the elements along any row or column by their corresponding cofactors and then summing the products. In this case, we can use the first row to calculate the determinant. Let's call this determinant D.

We can see that the first row has only one non-zero term, cosx. We can use this to our advantage by expanding the determinant along the first row. This means that we will multiply the element cosx by its corresponding cofactor and then add it to the determinant of the remaining submatrix (formed by removing the first row and column).

So, we have:

D = cosx * cofactor + determinant of submatrix

Now, let's look at the cofactor. The cofactor of cosx is (-1)^2 * determinant of the submatrix formed by removing the first row and column. We can see that this submatrix is exactly the same as the original matrix, just with a smaller size (n-1 rows and columns). So, the cofactor is (-1)^2 * D.

Substituting this into our original equation, we get:

D = cosx * (-1)^2 * D + determinant of submatrix

Simplifying, we get:

D = cosx * D + determinant of submatrix

Now, let's focus on the determinant of the submatrix. We can see that it is the same as the original matrix, just with a smaller size (n-1 rows and columns). So, we can use the same process again to evaluate this determinant. Expanding along the first row, we get:

determinant of submatrix = cosx * cofactor + determinant of sub-submatrix

Substituting this into our original equation, we get:

D = cosx * D + cosx * cofactor + determinant of sub-submatrix

We can continue this process until we reach the determinant of a 2x2 matrix, which is simply (2cosx)^2 - 1. Substituting this back into our equation, we get:

D = cosx * D + cosx * (-1)^n * (2cosx)^2 - 1

Simplifying, we get:

D = cosx * D + (-1)^n * (4cos^2x - 1)

Now, we can solve for D by rearranging the equation:

D - cosx * D = (-1)^