1.

Sin 20°.sin40°.sin80° = √3/8​

Answer»

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}


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