the question given below is of matrix :-

If A=[cos2theta sin2theta]

[-sin2theta cos2theta] find A^2

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Solution

It is a mistake
it is cos 4 theeta in last row
It is obtained by exoanding
cos 4 theeta=cos (2 theeta+2theeta)

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sin2θ + cos2θ	=	1.

Therefore is ROOT of sin2theta + ROOT of cos2theta must be equal to ROOT of one ?

Basically is sin ROOT theta +cos ROOT theta = 1

A B = 0

A = [\begin{matrix} {cos}^{2} θ & cos θ sin θ cos θ sin θ & {sin}^{2} θ \end{matrix}]

B = [\begin{matrix} {cos}^{2} ϕ & cos ϕ sin ϕ cos ϕ sin ϕ & {sin}^{2} ϕ \end{matrix}]

θ - ϕ

A = [\begin{matrix} {c o s}^{2} θ & c o s θ s i n θ c o s θ s i n θ & {s i n}^{2} θ \end{matrix}]

B = [\begin{matrix} {c o s}^{2} ϕ & c o s ϕ s i n ϕ c o s ϕ s i n ϕ & {s i n}^{2} ϕ \end{matrix}]

| θ - ϕ |

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin θ & cos θ - sin θ & - x & 1 cos θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin 2 θ & cos 2 θ - sin 2 θ & - x & 1 cos 2 θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

x \neq 0

θ \in (0, \frac{π}{2})

[\begin{matrix} cos 2 θ sin 2 θ \end{matrix} \begin{matrix} - sin 2 θ cos 2 θ \end{matrix}]

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