Sin 20-sin 40-sin 60-sin 80-MNR 1976- 81-

The correct option is C 3/16 sin 20^osin 40^∘sin 60^∘sin 80^∘ =1 2sin 20^∘sin 60^∘(2sin 40^∘sin 80^∘) =1 2sin 20^∘sin 60^∘(cos 40^∘ cos 120^∘) =1 2.√(3) 2sin 2 ...

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Byju's Answer

Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

sin 20∘sin 40...

Question

$s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$ =

[MNR 1976, 81]

-3/16

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5/16

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3/16

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-5/16

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Solution

The correct option is C
3/16

$s i n 20^{o} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$
= $\frac{1}{2} s i n 20^{\circ} s i n 60^{\circ} (2 s i n 40^{\circ} s i n 80^{\circ})$
= $\frac{1}{2} s i n 20^{\circ} s i n 60^{\circ} (c o s 40^{\circ} - c o s 120^{\circ})$
= $\frac{1}{2} . \frac{\sqrt{3}}{2} s i n 20^{\circ} (1 - 2 s i n^{2} 20^{\circ} + \frac{1}{2})$
= $\frac{\sqrt{3}}{2} s i n 20^{\circ} (\frac{3}{2} - 2 s i n^{2} 20^{\circ})$
= $\frac{\sqrt{3}}{8} (3 s i n 20^{\circ} - 4 s i n^{3} 20^{\circ})$
= $\frac{\sqrt{3}}{8} s i n 60^{\circ} = \frac{\sqrt{3}}{8} . \frac{\sqrt{3}}{2} = \frac{3}{16}$

Suggest Corrections

sin 20∘sin 40∘sin 60∘sin 80∘= [MNR 1976, 81]

The correct option is C 3/16 sin 20osin 40∘sin 60∘sin 80∘ =12sin 20∘sin 60∘(2sin 40∘sin 80∘) =12sin 20∘sin 60∘(cos 40∘−cos 120∘) =12.√32sin 20∘(1−2sin220∘+12) =√32sin 20∘(32−2sin220∘) =√38(3sin 20∘−4sin320∘) =√38sin60∘=√38.√32=316

$s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$ =

[MNR 1976, 81]