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Question

Let a=cosα+isinα then prove that an+1an=2cosnα for n being natural number.

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Solution

Given, a=cosα+isinα.
Now, an=cosnα+isinnα [ Using De Moivre's theorem] [ for n being natural number]
Then 1an=an=cosnαsinnα. [ Using De Moivre's theorem]
So an+1an=(cosnα+isinnα)+(cosnαisinnα)=2cosnα.

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