(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.
7 comments:
Show the IIt way to do that please
How 1/4(2 sin π/8...........) came..??
sin ^2 π/8 cos ^2 π/8
= 1/4( 4 sin ^2 π/8 cos ^2 π/8)
= 1/4( 2 sin π/8 cos π/8)^2
Can you prove this sum
(1+ sinπ/8)(1+ sin5π/8)(1+ sin9π/8)(1+ sin13π/8)=1/8
He multiplied and divided, (cos^2 π/8.sin^2 π/8) by 4 .
To get above result then he took root
but he multiplied 2 above and divided by 4 How?
No. I multiplied by 4 as 2 is inside square
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