Sin 6-sin 42-sin 66-sin 78-1-16

L.H.S. = sin 6^∘ sin 42^∘ sin 66^∘ sin 78^∘ =1/4 (2 sin 6^∘. sin60^∘)(2 sin 42^∘ . sin 78^∘) =1/4 (cos 60 cos 72^∘)(cos 36^∘ cos 120^∘) =1/4 (1/2 sin 18^∘ ) ...

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Byju's Answer

Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

sin 6∘sin 42∘...

Question

$s i n 6^{\circ} s i n 42^{\circ} s i n 66^{\circ} s i n 78^{\circ} = \frac{1}{16}$

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Solution

$L . H . S . = s i n 6^{\circ} s i n 42^{\circ} s i n 66^{\circ} s i n 78^{\circ} = \frac{1}{4} (2 s i n 6^{\circ} . s i n 60^{\circ}) (2 s i n 42^{\circ} . s i n 78^{\circ}) = \frac{1}{4} (c o s 60 - c o s 72^{\circ}) (c o s 36^{\circ} - c o s 120^{\circ}) = \frac{1}{4} (\frac{1}{2} - s i n 18^{\circ}) (\frac{\sqrt{5} + 1}{4} + \frac{1}{2}) = \frac{1}{4} (\frac{1}{2} - \frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4} + \frac{1}{2})$

$= \frac{1}{4} (\frac{2 - \sqrt{5} + 1}{4}) (\frac{\sqrt{5} + 1 + 2}{4}) = \frac{1}{4} (3^{2} - {\sqrt{5}}^{2}) = \frac{1}{64} (9 - 5) = \frac{1}{16}$
= RHS.

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Similar questions

Q. Prove that:

sin 6^{\circ} sin 42^{\circ} sin 66^{\circ} sin 78^{\circ} = \frac{1}{16}

Q. The value of

sin 78^{\circ} - sin 66^{\circ} - sin 42^{\circ} + sin 6^{\circ}

Question 1- sin 6 degree * sin 42 degree * sin 66 degree * sin 78 degree = ?

Question 2 - cos 6 degree* cos 42 degree * cos 66 degree * cos 78 degree =?

Q. Prove that:

\sin 6 ° \sin 42 ° \sin 66 ° \sin 78 ° = \frac{1}{16}

Q. The value of sin 78° − sin 66° − sin 42° + sin 60° is
(a)

\frac{1}{2}

(b)

- \frac{1}{2}

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sin 6∘ sin 42∘ sin 66∘ sin 78∘=116

L.H.S.=sin 6∘ sin 42∘ sin 66∘ sin 78∘=14 (2 sin 6∘. sin60∘)(2 sin 42∘ . sin 78∘)=14 (cos 60−cos 72∘)(cos 36∘−cos 120∘)=14 (12−sin 18∘)(√5+14+12)=14 (12−√5−14)(√5+14+12) =14(2−√5+14)(√5+1+24)=14 (32−√52)=164(9−5)=116 = RHS.

$s i n 6^{\circ} s i n 42^{\circ} s i n 66^{\circ} s i n 78^{\circ} = \frac{1}{16}$