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Exercise 7.4 · Q5

Q.If cos⁡2θ=0\cos 2\theta = 0, determine ∣0cos⁡θsin⁡θcos⁡θsin⁡θ0sin⁡θ0cos⁡θ∣2\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2.

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Expand the determinant to −(cos⁡3θ+sin⁡3θ)-(\cos^3\theta+\sin^3\theta); since cos⁡2θ=0\cos2\theta=0 gives θ=45∘\theta=45^\circ, this is −12-\tfrac1{\sqrt2}, whose square is 12\tfrac12.

Note the outer exponent: we must evaluate the square of the determinant.

Step 1 — expand the determinant. Let c=cos⁡θ, s=sin⁡θc=\cos\theta,\ s=\sin\theta and expand ∣0cscs0s0c∣\begin{vmatrix}0&c&s\\c&s&0\\s&0&c\end{vmatrix} along the first row:

D=0 (s⋅c−0)−c (c⋅c−0⋅s)+s (c⋅0−s⋅s)=−c⋅c2+s⋅(−s2)=−(c3+s3).D = 0\,(s\cdot c-0) - c\,(c\cdot c-0\cdot s) + s\,(c\cdot0-s\cdot s) = -c\cdot c^2 + s\cdot(-s^2) = -(c^3+s^3).

So D=−(cos⁡3θ+sin⁡3θ)D = -(\cos^3\theta+\sin^3\theta).

Step 2 — use the condition cos⁡2θ=0\cos2\theta=0. This gives 2θ=90∘2\theta = 90^\circ, i.e. θ=45∘\theta = 45^\circ, so cos⁡θ=sin⁡θ=12\cos\theta=\sin\theta=\tfrac1{\sqrt2}.

Step 3 — substitute. …

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