Question

If C = 2 cos q then the value of the determinant

D =$\left| \begin{matrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{matrix} \right|$is

• A. $\frac{\sin 4\theta}{\sin\theta}$
• B. $\frac{2\sin^{2}2\theta}{\sin\theta}$
• C. 4cos² q
• D. None

Answer 1

Answer: (D)

None

Answer 2

Given that C = 2cos q

D =$\left| \begin{matrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{matrix} \right|$= C(C² –1) –1(C –6)

D = 2 cos q (4 cos²q –1) – (2 cos q – 6)

D = 8 cos³ q – 4 cos q + 6