Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$ I am unable to show - without expanding, by using determinant properties - that 
$$\det\begin{bmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{bmatrix}=0$$
I am using trigonometric identities to solve this but I don't understand what would be the next step.
 A: Here we go: directly
$$\begin{align}\begin{vmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{vmatrix}&=\\&=
[1+2\cos a\cos b\cos(a+b)]-[\cos^2b+\cos^2a+\cos^2(a+b)]\\&=
1+\cos(a+b)\cdot [2\cos a\cos b-\cos (a+b)]-\cos^2a-\cos^2b\\&=
1+\cos(a+b)\cos(a-b)-\cos^2a-\cos^2b\\&=
1+\frac12[\cos(2a)+\cos(2b)]-\cos^2a-\cos^2b\\&=
1+\frac12[2\cos^2a-1+2\cos^2b-1]-\cos^2a-\cos^2b\\&=0\end{align}$$
Another method: triangulation
$$\begin{vmatrix}  1                       &\cos a               &\cos b\\
\cos a                     &1                &\cos(a+b) \\ 
\cos b                  &\cos(a+b)           &1
\end{vmatrix}\stackrel{(-\cos a)R_1+R_2\to R_2; \\(-\cos b)R_1+R_3\to R_3}{=}\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &1-\cos^2a                &\cos(a+b)-\cos a\cos b \\ 
0                  &\cos(a+b)-\cos a\cos b           &1-\cos^2b
\end{vmatrix}=\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &\sin^2a                &-\sin a\sin b \\ 
0                  &-\sin a\sin b           &\sin^2b
\end{vmatrix}\stackrel{\frac{\sin b}{\sin a}\cdot R_2+R_3\to R_3}{=}\\
\begin{vmatrix}  1                       &\cos a               &\cos b\\
0                     &\sin^2a                &-\sin a\sin b \\ 
0                  &0           &0
\end{vmatrix}=0.$$
A: This determinant is clearly zero.
The matrix
is the Gram matrix of the three unit vectors $(1,0)$, $(\cos a,\sin a)$
and $(\cos b,-\sin b)$ in the plane. Your matrix equals $AA^T$ where
$$A=\pmatrix{1&0\\\cos a&\sin a\\\cos b&-\sin b}$$
and so it is singular.
A: To flesh out my comment from earlier,$$\Delta:=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
\cos b & \cos\left(a+b\right) & 1
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & 0\\
\cos a & 1 & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$We may as well also simplify the second row/column:
$$\Delta=\left|\begin{array}{ccc}
1 & 0 & 0\\
\cos a & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left|\begin{array}{ccc}
1 & 0 & 0\\
0 & \sin^{2}a & -\sin a\sin b\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|.$$Well, now we just have an easy $2\times 2$ determinant.
A: Multiply the given matrix by
$$\begin{bmatrix}\sin (a+b)\\-\sin b \\ -\sin a             \end{bmatrix}$$
When you carry out the matrix multiplication, the identities for the sine of a sum and sine of a difference render all components of the product vector $=0$.
This of course fails when $a$ and $b$ are multiples of $\pi$, die to the proposed eigenvector being zero; but then the matrix trivially has all rows proportional to each other.  The breakdown of this eigenvector argument correlates with the null space becoming two-dimensional so that the zero-eigenvalue eigenvector is nonunique. 
