If A=aijn×n and f is a function, we define f(A)=faijn×n. Let A=π/2−θθ−θπ/2−θ. Then

sin⁡A=cos⁡θsin⁡θ−sin⁡θcos⁡θ and cos⁡A=sin⁡θcos⁡θcos⁡θsin⁡θ∴ |sin⁡A|=cos2⁡θ+sin2⁡θ=1Hence, sin A is invertible.Also, (sin⁡A)×(sin⁡A)T=cos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡θ−sin⁡θsin⁡θcos⁡θ=cos2⁡θ+sin2⁡θ00cos2⁡θ+sin2⁡θ=1001=IHence, sin A is orthogonal. Also,2sin⁡Acos⁡A=2cos⁡θsin⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θcos⁡θsin⁡θ=22sin⁡θcos⁡θcos2⁡θ+sin2⁡θcos2⁡θ−sin2⁡θ0=2sin⁡2θ1cos⁡2θ0≠sin⁡2A