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If A=(cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ)A = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}, then A2=A^2 =

Solution

✅ Correct Option: 1

Compute A2A^2 entry-wise:

(1,1):cos⁡22θ−sin⁡22θ=cos⁡4θ(1,1):\cos^2 2\theta-\sin^2 2\theta=\cos 4\theta.

(1,2):2sin⁡2θcos⁡2θ=sin⁡4θ(1,2):2\sin 2\theta\cos 2\theta=\sin 4\theta.

(2,1):−2sin⁡2θcos⁡2θ=−sin⁡4θ(2,1):-2\sin 2\theta\cos 2\theta=-\sin 4\theta.

(2,2):cos⁡4θ(2,2):\cos 4\theta. Result matches option 1.

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