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Appendix 1 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAppendix 13 Marks
Question
Show that if $A=\left[\begin{array}{ll} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right], \text { then } A^{n}=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]$
✓

Answer

we have
$P(n): A^{n}=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]$
we note that $\mathrm{P}(1): \mathrm{A}^{1}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
Therefore, p (1) is true.
Assume that p(k) is true i,e.
$P(k): A^{*}=\left[\begin{array}{ll} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]$
We want to prove that P (k +1) is true whenever p(k) is true ie
$P(k+1): A^{k+1}=\left[\begin{array}{ll} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]$
Now $A^{i+1}=A^{k} \cdot A$
Since p(k) is true, we have
$A^{\Delta 1}=\left[\begin{array}{rr} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$=\left[\begin{array}{cc} \cos k \theta \cos \theta-\sin k \theta \sin \theta & \cos k \theta \sin \theta+\sin k \theta \cos \theta \\ -\sin k \theta \cos \theta-\cos k \theta \sin \theta & -\sin k \theta \sin \theta+\cos k \theta \cos \theta \end{array}\right]$
$=\left[\begin{array}{ll} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction).

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