Prove D=−4sinθ sin2θ sin3θ.

01 / Standard#Your turn

D=\begin{vmatrix}1&1&1\\\sin2\theta&\sin4\theta&\sin8\theta\\\cos2\theta&\cos4\theta&\cos8\theta\end{vmatrix}

Official paper · jm02-2023 · 5(a)(ii) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Subtract column 1 from columns 2 and 3.

Hint 2

Apply the identities from part (i) to the resulting differences.

Worked solution

Expand along the simplified first row.
$D=\begin{vmatrix}\sin4\theta-\sin2\theta&\sin8\theta-\sin2\theta\\\cos4\theta-\cos2\theta&\cos8\theta-\cos2\theta\end{vmatrix}$
Extract the common sine factors from the columns.
$D=4\sin\theta\sin3\theta\begin{vmatrix}\cos3\theta&\cos5\theta\\-\sin3\theta&-\sin5\theta\end{vmatrix}$
The remaining determinant is −sin(5θ−3θ).
$D=-4\sin\theta\sin3\theta\sin2\theta$

D=−4sinθ sin2θ sin3θ.

Checks and common pitfalls: Column subtraction leaves a determinant unchanged; swapping columns would change its sign.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Expand along the simplified first row.
$D=\begin{vmatrix}\sin4\theta-\sin2\theta&\sin8\theta-\sin2\theta\\\cos4\theta-\cos2\theta&\cos8\theta-\cos2\theta\end{vmatrix}$
Extract the common sine factors from the columns.
$D=4\sin\theta\sin3\theta\begin{vmatrix}\cos3\theta&\cos5\theta\\-\sin3\theta&-\sin5\theta\end{vmatrix}$
The remaining determinant is −sin(5θ−3θ).
$D=-4\sin\theta\sin3\theta\sin2\theta$

Think first. Reveal a hint when the class is ready.

2023 JM02