$\displaystyle \begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix}$$\displaystyle =cos3\vartheta 2. I'll write x for theta. If we expand along the third row, the determinant is -cos(x)+2cos(x)(2(cos(x))^2-1)=4(cos(x))^3-3cos(x)=cos(x)(4(cos(x))^2-3)=cos(x)(2(2(cos(x))^2-1)-1)=cos(x)(2cos(2x)-1)=2cos(2x)cos(x)-cos(x)=cos(2x+x)=cos(3x) 3. I want to solve this problem by using the properties of determinants. 4. Originally Posted by roshanhero I want to solve this problem by using the properties of determinants. The I suggest you do elementary row operations to get the determinant into upper triangular form and then apply the appropriate theorems to calculate the determinant. 5. Originally Posted by roshanhero Prove that \displaystyle \begin{vmatrix} cos\theta & 1 & 0 \\1 & 2cos\theta & 1 \\0 & 1 & 2cos\theta\end{vmatrix}$$\displaystyle =cos3\vartheta$
You can do it by makin' $\displaystyle R_1-R_2\cos\theta$ and then expand by minors the first column.