Question

prove that sin20° sin40° sin80° sin90° = √3/8​

Answer 1

Answer:

we can write it as

Step-by-step explanation:

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

we should know that sinx sin(60-x) sin(60+x) = 1/4sin3x so using this

=1/4sin(3*20)

=1/4sin60

=1/4*$\sqrt{3\\}$/2

=$\sqrt{3}$/8

answer

Answer 2

It is  proved that  $sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree =\frac{\sqrt{3} }{8}$.

Step-by-step explanation:

Given:

$sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree =\frac{\sqrt{3} }{8}$.

To Find:

It is to be proved that $sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree =\frac{\sqrt{3} }{8}$.

Solution:

As given -  $sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree =\frac{\sqrt{3} }{8}$.

$LHS=sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree$

         $=sin20\textdegree sin40\textdegree sin80\textdegree \times 1$

         $=\frac{1}{2} [2sin20\textdegree sin40\textdegree ]sin80\textdegree$

          $=\frac{1}{2} [cos(20\textdegree -40\textdegree)-cos(20\textdegree +40\textdegree) ]sin80\textdegree$

          $=\frac{1}{2} [cos20\textdegree-cos60\textdegree ]sin80\textdegree$

          $=\frac{1}{2} [cos20\textdegree-\frac{1}{2} ]sin80\textdegree$

         $=\frac{1}{4} [2cos20\textdegree-1 ]sin80\textdegree$

         $=\frac{1}{4} [2cos20\textdegree sin80\textdegree - sin80\textdegree]$

         $=\frac{1}{4} [sin(20\textdegree +80\textdegree)-sin(20\textdegree -80\textdegree) -sin80\textdegree]$

          $=\frac{1}{4} [sin100\textdegree -sin( -60\textdegree) -sin80\textdegree]$

          $=\frac{1}{4} [sin100\textdegree +sin60\textdegree -sin80\textdegree]$

          $=\frac{1}{4} [sin100\textdegree -sin80\textdegree +\frac{\sqrt{3} }{2} ]$

           $=\frac{1}{4} [2 cos \frac{(100\textdegree+80\textdegree )}{2} sin \frac{(100\textdegree-80\textdegree )}{2} +\frac{\sqrt{3} }{2} ]$

          $=\frac{1}{4} [2 cos90 \textdegree sin 10\textdegree +\frac{\sqrt{3} }{2} ]$

          $=\frac{1}{4} [2\times 0\times sin 10\textdegree +\frac{\sqrt{3} }{2} ]$

          $=\frac{1}{4} [0 +\frac{\sqrt{3} }{2}]$

         $=\frac{\sqrt{3} }{8}$

        $=RHS$

$LHS=RHS$

Hence, It is  proved that  $sin20\textdegree sin40\textdegree sin80\textdegree sin90\textdegree =\frac{\sqrt{3} }{8}$.

PROJECT CODE  #SPJ3