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Question
Prove the following:
`(1 + cos pi/8)(1 + cos (3pi)/8)(1 + cos (5pi)/8)(1 + cos (7pi)/8) = 1/8`
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Solution
L.H.S. = `(1 + cos pi/8)(1 + cos (3pi)/8)(1 + cos (5pi)/8)(1 + cos (7pi)/8)`
Since, cos (π – θ) = – cos θ
∴ `cos (7pi)/8 = cos(pi - pi/8) = -cos pi/8` ...(i)
and `cos (5pi)/8 = cos(pi - (3pi)/8) = -cos (3pi)/8` ...(ii)
∴ L.H.S. = `(1 + cos pi/8)(1 + cos (3pi)/8)*(1 - cos (3pi)/8)(1 - cos pi/8)` ...[From (i) and (ii)
= `(1 - cos^2 pi/8)(1 - cos^2 (3pi)/8)`
= `sin^2 pi/8 sin^2 (3pi)/8`
= `1/4(2sin pi/8sin (3pi)/8)^2`
= `1/4[cos(pi/8 - (3pi)/8)-cos(pi/8 + (3pi)/8)]^2`
= `1/4[cos(-pi/4)-cos(pi/2)]^2`
= `1/4(cos(pi/4) - 0)^2`
= `1/4(1/sqrt(2))^2`
= `1/4(1/2)`
= `1/8`
= R.H.S.
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