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Basic Inverse Trigonometric Functions

Let F α=cosα-...

Question

Let $F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] and G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}]$

Show that
(i) ${[F (α)]}^{- 1} = F (- α)$
(ii) ${[G (β)]}^{- 1} = G (- β)$
(iii) ${[F (α) G (β)]}^{- 1} = G (- β) F (- α)$ .

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Solution

$(i) F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow F (- α) = [\begin{matrix} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] Now, C_{11} = |\begin{matrix} \cos α & 0 \\ 0 & 1 \end{matrix}| = \cos α, C_{12} = - |\begin{matrix} \sin α & 0 \\ 0 & 1 \end{matrix}| = - \sin α and C_{13} = |\begin{matrix} \sin α & \cos α \\ 0 & 0 \end{matrix}| = 0 C_{21} = - |\begin{matrix} - \sin α & 0 \\ 0 & 1 \end{matrix}| = \sin α, C_{22} = |\begin{matrix} \cos α & 0 \\ 0 & 1 \end{matrix}| = \cos α and C_{23} = - |\begin{matrix} \cos α & - \sin α \\ 0 & 0 \end{matrix}| = 0 C_{31} = |\begin{matrix} - \sin α & 0 \\ \cos α & 0 \end{matrix}| = 0, C_{32} = - |\begin{matrix} \cos α & 0 \\ \sin α & 0 \end{matrix}| = 0 and C_{33} = |\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}| = 1 \Rightarrow adj \{F (α)\} = {[\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}]}^{T} = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow |F (α)| = 1 ∴ {[F (α)]}^{- 1} = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] . . . (1) \Rightarrow {[F (α)]}^{- 1} = F (- α) (ii) G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}] \Rightarrow G (- β) = [\begin{matrix} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{matrix}] = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 0 & \cos β \end{matrix}| = \cos β, C_{12} = - |\begin{matrix} 0 & 0 \\ - \sin β & \cos β \end{matrix}| = 0 and C_{13} = |\begin{matrix} 0 & 1 \\ - \sin β & 0 \end{matrix}| = \sin β C_{21} = - |\begin{matrix} 0 & \sin β \\ 0 & \cos β \end{matrix}| = 0, C_{22} = |\begin{matrix} \cos β & \sin β \\ - \sin β & \cos β \end{matrix}| = 1 and C_{23} = - |\begin{matrix} \cos β & 0 \\ - \sin β & 0 \end{matrix}| = 0 C_{31} = |\begin{matrix} 0 & \sin β \\ 1 & 0 \end{matrix}| = - \sin β, C_{32} = - |\begin{matrix} \cos β & \sin β \\ 0 & 0 \end{matrix}| = 0 and C_{33} = |\begin{matrix} \cos β & 0 \\ 0 & 1 \end{matrix}| = \cos β adj \{G (β)\} = {[\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}]}^{T} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] |G (β)| = 1 ∴ G (β)^{- 1} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] . . . (2) \Rightarrow G (β)^{- 1} = = G (- β) (iii) F (α) = [\begin{matrix} \cos α & - \sin α & 0 \\ \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow F (- α) = [\begin{matrix} \cos (- α) & - \sin (- α) & 0 \\ \sin (- α) & \cos (- α) & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] . . . (3) G (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}] \Rightarrow G (- β) = [\begin{matrix} \cos (- β) & 0 & \sin (- β) \\ 0 & 1 & 0 \\ - \sin (- β) & 0 & \cos (- β) \end{matrix}] = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] . . . (4) {[F (α) G (β)]}^{- 1} = {[G (β)]}^{- 1} {[F (α)]}^{- 1} = [\begin{matrix} \cos β & 0 & - \sin β \\ 0 & 1 & 0 \\ \sin β & 0 & \cos β \end{matrix}] [\begin{matrix} \cos α & \sin α & 0 \\ - \sin α & \cos α & 0 \\ 0 & 0 & 1 \end{matrix}] [Using eqn (1) and (2)] = G (- β) F (- α) [Using eqn (3) and (4)]$

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