Question Number 9 by user1 last updated on 25/Jan/15
![Let A= [(( 0),(−tan(x/2))),((tan(x/2)),( 0)) ] and I is the identity matrix of order 2. Show that (I+A)=(I−A)∙ [((cos x),(−sin x)),((sin x),( cos x)) ].](https://www.tinkutara.com/question/Q9.png)
$$\mathrm{Let}\:{A}=\begin{bmatrix}{\:\:\:\mathrm{0}}&{−\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}\\{\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}\:\mathrm{and}\:{I}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{identity}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{order}\:\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}\: \\ $$$$\left({I}+{A}\right)=\left({I}−{A}\right)\centerdot\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix}. \\ $$
Answered by user1 last updated on 30/Oct/14
![Let tan (x/2)=t Then, cos x=((1−tan^2 (x/2))/(1+tan^2 (x/2)))=((1−t^2 )/(1+t^2 )) and sin x=((2tan (x/2))/(1+tan^2 (x/2)))=((2t)/(1+t^2 )) ∴ (I+A)= [(1,0),(0,1) ]+ [(0,(−t)),(t,( 0)) ]= [(1,(−t)),(t,( 1)) ] (I−A)= [(1,0),(0,1) ]− [(0,(−t)),(t,( 0)) ]= [(( 1),t),((−t),1) ] ∴(I−A)∙ [((cos x),(−sin x)),((sin x),( cos x)) ] = [(( 1),t),((−t),1) ] [(((1−t^2 )/(1+t^2 )),((−2t)/(1+t^2 ))),(((2t)/(1+t^2 )),((1−t^2 )/(1+t^2 ))) ] = [(( ((1−t^2 )/(1+t^2 ))+((2t^2 )/(1+t^2 ))),( ((−2t)/(1+t^2 ))+((t(1−t^2 ))/(1+t^2 )))),((((−t(1−t^2 ))/(1+t^(2 ) ))+((2t)/(1+t^2 ))),( ((2t^2 )/(1+t^2 ))+((1−t^2 )/(1+t^(2 ) )))) ] = [(1,(−t)),(t,( 1)) ]=(I+A) Hence, (I+A)=(I−A) [((cos x),(−sin x)),((sin x),( cos x)) ]](https://www.tinkutara.com/question/Q10.png)
$$\mathrm{Let}\:\:\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}={t} \\ $$$$\mathrm{Then},\:\:\mathrm{cos}\:{x}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\:\:\mathrm{sin}\:{x}=\frac{\mathrm{2tan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\therefore\:\:\left({I}+{A}\right)=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}+\begin{bmatrix}{\mathrm{0}}&{−{t}}\\{{t}}&{\:\:\:\:\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{−{t}}\\{{t}}&{\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\:\:\:\left({I}−{A}\right)=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}−\begin{bmatrix}{\mathrm{0}}&{−{t}}\\{{t}}&{\:\:\:\:\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\mathrm{1}}&{{t}}\\{−{t}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\therefore\left({I}−{A}\right)\centerdot\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\:\:\:\:\mathrm{1}}&{{t}}\\{−{t}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}&{\frac{−\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\\{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}&{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}&{\:\frac{−\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }}\\{\frac{−{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}\:} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}&{\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}\:} }}\end{bmatrix} \\ $$$$=\begin{bmatrix}{\mathrm{1}}&{−{t}}\\{{t}}&{\:\:\:\:\mathrm{1}}\end{bmatrix}=\left({I}+{A}\right) \\ $$$$\mathrm{Hence},\:\:\left({I}+{A}\right)=\left({I}−{A}\right)\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix} \\ $$