Sin 36-sin 72-sin 108-sin 144-5-16

LHS, sin 36^∘ sin 72^∘ sin 108^∘ sin 144^∘ [∵ sin 144^∘ = sin (180^∘ 36^∘)=sin 36^∘ and sin 108^∘ = sin (180^∘ 78^∘) =sin 72^∘ ] = sin 36^∘. sin 72^∘. s ...

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Byju's Answer

Standard IX

Mathematics

Values of Trigonometric Ratios

sin 36∘sin 72...

Question

$s i n 36^{\circ} s i n 72^{\circ} s i n 108^{\circ} s i n 144^{\circ} = \frac{5}{16}$

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Solution

$L H S, s i n 36^{\circ} s i n 72^{\circ} s i n 108^{\circ} s i n 144^{\circ} [∵ s i n 144^{\circ} = s i n (180^{\circ} - 36^{\circ}) = s i n 36^{\circ} a n d s i n 108^{\circ} = s i n (180^{\circ} - 78^{\circ}) = s i n 72^{\circ}] = s i n 36^{\circ} . s i n 72^{\circ} . s i n 72^{\circ} . s i n 36^{\circ} = \frac{1}{4} (2 s i n 36^{\circ} . s i n 72^{\circ})^{2} = \frac{1}{4} (2 s i n 36^{\circ} c o s 18^{\circ})^{2} [∵ s i n 72^{\circ} = c o s 18^{\circ}]$

$= \frac{4}{4} {(\frac{\sqrt{10 - 2 \sqrt{5}}}{4} . \frac{\sqrt{10 + 2 \sqrt{5}}}{4})}^{2} = \frac{1}{64} (10 - 2 \sqrt{5}) (10 + 2 \sqrt{5}) = \frac{100 - 20}{64 \times 4} = \frac{80}{256} = \frac{5}{16}$
= RHS.

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Similar questions

sin 36^{\circ} sin 72^{\circ} . sin 108^{\circ} . sin 144^{\circ} = \frac{k}{16} .

Find the value of

k

Simplify:

sin 36 . sin 27 . sin 108 . sin 144 = \frac{5}{16}

Q. Prove that:

\sin 36 ° \sin 72 ° \sin 108 ° \sin 144 ° = \frac{5}{16}

s i n 36^{\circ} s i n 72^{o} s i n 108^{\circ} s i n 144^{\circ}

Q. Without using trigonometric tables, prove that:

(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) sin35° sin55° − cos35° cos55° = 0
(v) (sin72° + cos18°)(sin72° − cos18°) = 0
(vi) tan48° tan23° tan42° tan67° = 1

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sin 36∘ sin 72∘ sin 108∘ sin 144∘=516

$s i n 36^{\circ} s i n 72^{\circ} s i n 108^{\circ} s i n 144^{\circ} = \frac{5}{16}$