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Question

For any integer k, If ak=cos(kπ7)+isin(kπ7), then value of the expression 12k=1|ak+1ak|3k=1|a4k1a4k2| is

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Solution

ak=cos(kπ7)+isin(kπ7)=ei(kπ7)
So,
ak+1ak=ei((k+1)π7)ei(kπ7)=ei(kπ7)(eiπ71)=ei(kπ7)eiπ14(eiπ14eiπ14)=ei((2k+1)π7)×[2isin(π14)]|ak+1ak|=2sin(π14)
Replacing k with 4k2 in the above expression,
|a4k1a4k2|=2sin(π14)
Hence the value of the expression,
12k=1|a(k+1)ak|3k=1|a(4k1)a(4k2)|=246=4


Alternate Solution
ak=cos(kπ7)+isin(kπ7)
|ak+1ak|=∣ ∣(cos(k+1)π7coskπ7)+i(sin(k+1)π7sinkπ7)∣ ∣
= 22(cos(k+1)π7coskπ7+sin(k+1)π7sinkπ7)
|ak+1ak|=22cosπ7
=2(1cosπ7)
=2sinπ14
Replace k with 4k2 in above equation
|a4k2a4k2|=2sinπ14
Hence the value of the expression,
12k=1|a(k+1)ak|3k=1|a(4k1)a(4k2)|=246=4

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