Question

Find the value of sin 20° sin 40° sin 80°​

Answer 1

Given to find the value of :-

sin20° . sin40° . sin80°

Formula to know :-

${sin\theta \times sin(\alpha-\theta) \times sin(\alpha+\theta) = \dfrac{1}{4}sin3\theta}$

SOLUTION:-

sin20° × sin40° × sin80°

sin40° can be written as sin(60°-20°)

sin80° can be written as sin(60°+20°)

= sin20° × sin(60°-20°) × sin (60° +20°)

It is in form of ${sin\theta \times sin(\alpha-\theta) \times sin(\alpha+\theta) = \dfrac{1}{4}sin3\theta}$

Since here ${\theta = 20^{\circ}}$

${sin20^{\circ} \times sin40^{\circ} \times sin80^{\circ} = \dfrac{1}{4} sin3(20^{\circ}})$

${sin20^{\circ} \times sin40^{\circ} \times sin80^{\circ} = \dfrac{1}{4} sin(60^{\circ}})$

Also we know that sin60° = √3/2

${=} \dfrac{1}{4} \times \dfrac{\sqrt{3}}{2}$

${=} \dfrac{\sqrt{3}}{8}$

So, the value of sin 20° sin 40° sin 80° = √3/8

Know more :-

${cos\theta \times cos(\alpha-\theta) \times cos(\alpha+\theta) = \dfrac{1}{4}cos3\theta}$

${tan\theta \times tan(\alpha-\theta) \times tan(\alpha+\theta) = tan3\theta}$

${cot\theta \times cot(\alpha-\theta) \times cot(\alpha+\theta) = cot3\theta}$

${sin^2\theta \times sin^2(\alpha-\theta) \times sin^2(\alpha+\theta) = \dfrac{3}{2}}$

${cos^2\theta \times cos^2(\alpha-\theta) \times cos^2(\alpha+\theta) = \dfrac{3}{2}}$