Matrix and trigonometry

Algebra Level 5

( 3 + 1 2 2 3 1 2 2 1 3 2 2 1 + 3 2 2 ) 2017 = ( 6 2 4 2 + 6 4 2 6 4 6 2 4 ) k \large \begin{pmatrix} \frac{\sqrt{3}+1}{2\sqrt{2}} & \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \frac{1-\sqrt{3}}{2\sqrt{2}} & \frac{1+\sqrt{3}}{2\sqrt{2}} \end{pmatrix}^{2017} = \begin{pmatrix} \frac{\sqrt{6}-\sqrt{2}}{4} & \frac{\sqrt{2}+\sqrt{6}}{4} \\ \frac{-\sqrt{2}-\sqrt{6}}{4} & \frac{\sqrt{6}-\sqrt{2}}{4} \end{pmatrix}^{k} If k k satisfy the equation above and k k is of the form a + b n a+bn where a a and b b are positive intergers for n = 0 , 1 , 2 , n =0, 1 , 2 , \cdots , find a + b a+b .


The answer is 29.

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1 solution

Tommy Li
Aug 26, 2017

( cos θ sin θ sin θ cos θ ) n = ( cos n θ sin n θ sin n θ cos n θ ) \large \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}^n = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}

( 3 + 1 2 2 3 1 2 2 1 3 2 2 1 + 3 2 2 ) 2017 = ( cos 1 5 o sin 1 5 o sin 1 5 o cos 1 5 o ) 2017 = ( cos ( 2017 15 ) o sin ( 2017 15 ) o sin ( 2017 15 ) o cos ( 2017 15 ) o ) = ( cos 1 5 o sin 1 5 o sin 1 5 o cos 1 5 o ) \large \begin{pmatrix} \frac{\sqrt{3}+1}{2\sqrt{2}} & \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \frac{1-\sqrt{3}}{2\sqrt{2}} & \frac{1+\sqrt{3}}{2\sqrt{2}} \end{pmatrix}^{2017} = \begin{pmatrix} \cos15^o & \sin15^o \\ -\sin15^o & \cos15^o \end{pmatrix}^{2017} = \begin{pmatrix} \cos(2017\cdot15)^o & \sin(2017\cdot15)^o \\ -\sin(2017\cdot15)^o & \cos(2017\cdot15)^o \end{pmatrix} = \begin{pmatrix} \cos15^o & \sin15^o \\ -\sin15^o & \cos15^o \end{pmatrix}

( 6 2 4 2 + 6 4 2 6 4 6 2 4 ) k = ( cos 7 5 o sin 7 5 o sin 7 5 o cos 7 5 o ) k = ( cos 75 k o sin 75 k o sin 75 k o cos 75 k o ) \large \begin{pmatrix} \frac{\sqrt{6}-\sqrt{2}}{4} & \frac{\sqrt{2}+\sqrt{6}}{4} \\ \frac{-\sqrt{2}-\sqrt{6}}{4} & \frac{\sqrt{6}-\sqrt{2}}{4} \end{pmatrix}^{k} = \begin{pmatrix} \cos75^o & \sin75^o \\ -\sin75^o & \cos75^o \end{pmatrix}^{k} = \begin{pmatrix} \cos75k^o & \sin75k^o \\ -\sin75k^o & \cos75k^o \end{pmatrix}

75 k 15 ( m o d 360 ) \large 75k \equiv 15 \pmod {360}

k = 5 + 24 n \large \Rightarrow k = 5+24n for n = 0 , 1 , 2 , \large n =0, 1 , 2 , \cdots

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