# Given x + x-1 = 2 cos y,          x2 + x-2 = 2 cos (2y),Prove that xn + x-n = 2 cos (ny) by mathematical Induction.

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
8 years ago
Dear student,
Let us assume
$x^{k}+x^{-k}=2cos\left ( ky \right )$
and thus
$x^{k-1}+x^{-k+1}=2cos\left ( \left (k-1 \right )y \right )$
So, we need to prove it for
$x^{k+1}+x^{-k-1}=2cos\left ( \left (k+1 \right )y \right )$
Proof$\left (x^{k}+x^{-k} \right )\times \left (x^{1}+x^{-1} \right )= 4cos\left ( 2ky)cos\left ( 2y \right ) \Rightarrow \left (x^{k+1}+x^{-k-1} \right )= 2\left ( cos\left ( 2k+1 \right )y \right +cos\left ( 2k-1 \right )y \right)-\left (x^{-k+1}+x^{k-1} \right )= 2\left ( cos\left ( 2k+1 \right )y \right \right)$:
$\left (x^{k}+x^{-k} \right )\times \left (x^{1}+x^{-1} \right )=4cos\left ( 2ky)cos\left ( 2y \right ) \Rightarrow \left (x^{k+1}+x^{-k-1} \right )=2\left ( cos\left ( 2k+1 \right )y \right +cos\left ( 2k-1 \right )y \right)-\left (x^{-k+1}+x^{k-1} \right )=2\left ( cos\left ( 2k+1 \right )y \right \right)$
Regards
Sumit