I need help with the underlined part. Thanks in advance
Let $A_n$ be the $n\times n$ matrix given by $$a_{ij}= \begin{cases} 0 & \text{if }|i-j|>1, \\ 1 & \text{if }|i-j|=1, \\ 2\cos\vartheta & \text{if }i=j. \end{cases} $$ If $\Delta_n=\det A$, prove that $$\Delta_{n+2}-2\cos\vartheta\Delta_{n+1}+\Delta_n=0.$$
Hence show by induction that, for $0<\vartheta<\pi$, $$\det A_n=\frac{\sin(n+1)\vartheta}{\sin\vartheta}.$$
It is easy to check this for $n=1$ and $n=2$.
In the inductive step you simply have to apply $$\frac{\sin\alpha+\sin\beta}2=\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2$$ to $\alpha=(n+3)\vartheta$ and $\beta=(n+1)\vartheta$.
(This is one of the sum-to-product identities.)
Adding more details:
The inductive step (if we assume the validity of the formula for $n$ and $n+1$ and we want to verify whether it is true for $n+2$) is equivalent to verifying whether $\Delta_{n+2}-2\cos\vartheta\Delta_{n+1}+\Delta_n\overset?=0$ holds for $\Delta_n=\frac{\sin(n+1)\vartheta}{\sin\vartheta}$.
This is equivalent to
$\Delta_{n+2}+\Delta_n\overset?=2\cos\vartheta\Delta_{n+1}$
$\frac{\sin(n+3)\vartheta}{\sin\vartheta}+\frac{\sin(n+1)\vartheta}{\sin\vartheta}\overset?=\cos\vartheta\cdot\frac{\sin(n+2)\vartheta}{\sin\vartheta}$
$\sin(n+3)\vartheta+\sin(n+1)\vartheta\overset?=\cos\vartheta\sin(n+2)\vartheta$
This is precisely the above equality with $\alpha=(n+3)\vartheta$ and $\beta=(n+1)\vartheta$.