θ$$=2$$$$π$$$$2009$$ then $$cos$$θ$$cos2$$θ$$cos3$$θ……..$$cos1004$$θ

Question

θ$$=2$$$$π$$$$2009$$ then $$cos$$θ$$cos2$$θ$$cos3$$θ……..$$cos1004$$θ

Infinity Learn · Accepted Answer

We have θ$$=2$$$$π$$$$2009$$⇒$$2009$$θ$$=2$$π⇒θ$$=2$$π$$-2008$$θ⇒$$cos$$θ$$=cos2008$$θ    $$cos2$$θ$$=cos2007$$θ and so onAlso $$sin$$θ$$=$$$$sin$$(2$$π$$-2008$$θ$$)=-sin2008$$θ  and so on. Let $$P=$$$$cos$$θ$$cos2$$θ$$cos3$$θ…………$$cos1004$$θand $$Q=$$$$sin$$θ$$sin2$$θ$$sin4$$θ…………$$sin1004$$θNow $$21004PQ=sin2$$θ$$sin4$$θ……………$$sin2008$$θ →$$(1) Since $$sin$$θ$$=-sin2008$$θ          $$sin3$$θ$$=-sin2006$$θ          ............................    $$sin1003$$θ$$=-sin1006$$θ∴ From (1) we have $$21004PQ=$$$$sin$$θ$$sin2$$θ$$sin3$$θ$$sin4$$θ………….$$sin1004$$θ                                                 $$=Q$$⇒$$P=121004$$

$θ = \frac{2 π}{2009} then \cos θ \cos 2 θ \cos 3 θ \dots \dots . . \cos 1004 θ$

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θ=2π2009 then cosθcos2θcos3θ……..cos1004θ

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$θ = \frac{2 π}{2009} then \cos θ \cos 2 θ \cos 3 θ \dots \dots . . \cos 1004 θ$