Question

sin 20°.sin40°.sin80° = √3/8​

Answer 1

Answer:

Consider LHS

sin 20 × sin 40 × sin60 × sin80

Sin Values

sin 0° = √(0/4) = 0

sin 30° = √(1/4) = ½

sin 45° = √(2/4) = 1/√2

sin 60° = √3/4 = √3/2

sin 90° = √(4/4) = 1

LHS = sin60 [sin20 × sin40 × sin80]

LHS = √3/2[sin20 × sin(60 – 20) × sin(60 + 20)]

LHS = √3/2[sin 3(20)/4]

LHS = √3/2[sin 60/4]

LHS = √3/2[√3/2 × 4]

LHS = √3/2 × √3/8

LHS = 3/16

∴ LHS = RHS

Answer 2

Step-by-step explanation:

To Prove :-

sin20°.sin40°.sin80° = $\sf\dfrac{\sqrt{3}}{8}$

Solution :-

Taking L.H.S :-

sin20°.sin40°.sin80°

Multiplying and Dividing with '2' :-

= $\sf\dfrac{2}{2}\times$ sin20°.sin40°.sin80°

=

$\dfrac{1}{2}(2 \times sin20 \degree \times sin40 \degree \times (sin80 \degree))$

We know that :-

cos(A-B)-cos(A+B) = 2sinAsinB

$= \dfrac{1}{2}((cos(40 \degree - 20 \degree) - cos(40 \degree + 20 \degree) ) \times sin80 \degree)$

$= \dfrac{1}{2} ((cos20 \degree - cos60 \degree) \times sin80 \degree)$

$= \dfrac{1}{2} ((cos20 \degree - \dfrac{1}{2} ) \times sin80 \degree)$

$= \dfrac{1}{2} \bigg( \dfrac{2cos20 \degree - 1}{2} \bigg) \times sin80 \degree$

$= \dfrac{1}{4}((2cos20 \degree - 1) \times sin80 \degree)$

$= \dfrac{1}{4}((2cos20 \degree \times sin80 \degree ) - sin80 \degree)$

we know that :-

sin(A+B)-sin(A-B) = 2cosAsinB

$= \dfrac{1}{4}(sin100 \degree + sin60 \degree - sin80 \degree )$

$\dfrac{1}{4}(sin100 \degree - sin80 \degree + sin60 \degree )$

$= \dfrac{1}{4}((2cos90 \degree \times sin20 \degree) + \dfrac{ \sqrt{3} }{2} )$

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

Since , cos90° = 0

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