If `theta=(2pi)/(2009), then costhetacos 2thetacos3theta... cos1004theta` is

Options

A. 0
B. `(1)/(2^(2008))`
C. `(1)/(2^(1004))`
D. `-(1)/(2^(1004))`

Answer

C

Explanation

Let `P=costhetacos2thetacos 3theta...sin1004theta`
Then, `2^(1004)PQ=sin2theta sin4thetasin6theta...sin2008 theta`
`implies2^(1004)PQ=(sin2thetasin4thetasin60...sin2008theta)(sin1006thetasin1008theta...sin2008theta)`
`2^(1004)PQ=(sin2thetasin4thetasin6theta...sin1004theta)`
`{sin(2pi-1003theta)sin(2pi-1001theta)...sin(2pi-theta)}`
`implies2^(1004)PQ=(-1)^(502)sinthetasin2thetasin3thetasin4theta.....sin1003thetasin1001theta`
`implies2^(1004)PQ=QimpliesP=(1)/(2^(1004))`