1. ## Determinant

Prove that
$\begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix}$ $=cos3\vartheta$

2. I'll write x for theta.
If we expand along the third row, the determinant is
-cos(x)+2cos(x)(2(cos(x))^2-1)=4(cos(x))^3-3cos(x)=cos(x)(4(cos(x))^2-3)=cos(x)(2(2(cos(x))^2-1)-1)=cos(x)(2cos(2x)-1)=2cos(2x)cos(x)-cos(x)=cos(2x+x)=cos(3x)

3. I want to solve this problem by using the properties of determinants.

4. Originally Posted by roshanhero
I want to solve this problem by using the properties of determinants.
The I suggest you do elementary row operations to get the determinant into upper triangular form and then apply the appropriate theorems to calculate the determinant.

5. Originally Posted by roshanhero
Prove that
$\begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix}$ $=cos3\vartheta$
You can do it by makin' $R_1-R_2\cos\theta$ and then expand by minors the first column.

Originally Posted by roshanhero
I want to solve this problem by using the properties of determinants.
If you want to solve it, do it, show some progress but you're gonna have to change the way on asking your questions.