First slide

Multiple and sub- multiple Angles

Question

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

Very difficult

A

0
B

$\frac{1}{2^{2008}}$
C

$\frac{1}{2^{1004}}$
D

$10$

Solution

$\begin{array}{l} W e h a v e θ = \frac{2 π}{2009} \Rightarrow 2009 θ = 2 π \\ \Rightarrow θ = 2 π - 2008 θ \\ \Rightarrow \cos θ = \cos 2008 θ \\ \cos 2 θ = \cos 2007 θ a n d s o o n \\ A l s o \sin θ = \sin (2 π - 2008 θ) = - \sin 2008 θ a n d s o o n . \\ Let P = \cos θ \cos 2 θ \cos 3 θ \dots \dots \dots \dots \cos 1004 θ \\ a n d Q = \sin θ \sin 2 θ \sin 4 θ \dots \dots \dots \dots \sin 1004 θ \end{array}$
$\begin{array}{l} N o w 2^{1004} P Q = \sin 2 θ \sin 4 θ \dots \dots \dots \dots \dots \sin 2008 θ \to (1) \\ Since \sin θ = - \sin 2008 θ \\ \sin 3 θ = - \sin 2006 θ \\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . \\ \sin 1003 θ = - \sin 1006 θ \\ ∴ From (1) we have 2^{1004} P Q = \sin θ \sin 2 θ \sin 3 θ \sin 4 θ \dots \dots \dots \dots . \sin 1004 θ \\ = Q \\ \Rightarrow P = \frac{1}{2^{1004}} \end{array}$

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