Prove that sin 20 - sin40 - sin60 - sin80 - 3-16- - JEE Maths Q-A

Consider LHS sin 20 ×sin 40 ×sin60 ×sin80 0ex0ex= sin60 [sin20 ×sin40 ×sin80] 0ex0ex= √(3)/2[sin20 ×sin(60 – 20) ×sin(60 + 20)] 0ex0ex= √(3)/2[sin 3(20)/4] ...

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Byju's Answer

Standard XII

Physics

Chain Rule of Differentiation

Prove that si...

Question

Prove that $\sin 20 \times \sin 40 \times \sin 60 \times \sin 80 = \frac{3}{16}$

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Solution

Consider LHS
$\sin 20 \times \sin 40 \times \sin 60 \times \sin 80 = \sin 60 [\sin 20 \times \sin 40 \times \sin 80] = \frac{\sqrt{3}}{2} [\sin 20 \times \sin (60 - 20) \times \sin (60 + 20)] = \frac{\sqrt{3}}{2} [\frac{\sin 3 (20)}{4}] [∵ \sin A \sin (60 - A) \sin (60 + A) = \frac{1}{4} \sin 3 A] = \frac{\sqrt{3}}{2} [\frac{\sin 60}{4}] = \frac{\sqrt{3}}{2} [\frac{\sqrt{3}}{2} \times 4] = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{8} = \frac{3}{16}$
$= RHS ∴ LHS = RHS$
Hence proved

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Prove that sin20×sin40×sin60×sin80=316

Consider LHS sin20×sin40×sin60×sin80=sin60[sin20×sin40×sin80]=32[sin20×sin(60–20)×sin(60+20)]=32sin3(20)4∵sinAsin(60−A)sin(60+A)=14sin3A=32sin604=3232×4=32×38=316=RHS∴LHS=RHSHence proved

Prove that $\sin 20 \times \sin 40 \times \sin 60 \times \sin 80 = \frac{3}{16}$