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prove that sin20 sin40 sin60 sin80=3/16
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sin 20 .sin 40.sin60.sin 80 = 3/2 . sin 20 .sin (60 - 20) sin (60+20) = 3/2 . sin 20.(sin 2 60 - sin 2 20) = 3/2 .sin 20 . (3/4 - sin 2 20) = 3/8 .(3 sin 20 - 4 sin 3 20) = 3/8 .(sin 3 . 20) = 3/8 sin 60 = 3/16
\[sum_{1/sqrt{n ^{2}+n}?}^{?}\] show that the sum given is 1
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