Question

prove that sin∅.sin(60°-∅).sin(60°+∅)=1/4 sin3∅​

Answer 1

Step-by-step explanation:

see the attachment mate

Answer 2

SOLUTION

L.H.S

$= > sin \theta.sin(60 - \theta).sin(60 + \theta) \\ = > sin \theta.(sin60cos \theta - cos60 \: sin \theta).(sin60 \cos \theta + cos60 \sin \theta) \\ = > sin \theta.(sin {}^{2} 60 \cos {}^{2} \theta - {cos}^{2} 60 {sin}^{2} \theta) \\ = > sin \theta.( \frac{3}{4} ) {cos}^{2} \theta - (\frac{1}{4} ) {sin}^{2} \theta \\ = > sin \theta.( \frac{3}{4} ) - ( \frac{3}{4} )sin {}^{2} \theta - ( \frac{1}{4} )sin {}^{2} \theta \\ = > sin \theta( \frac{3}{4} ) - sin { }^{2} \theta \\ = > ( \frac{3}{4} )sin \theta - sin {}^{3} \theta \\ = > \frac{1}{4} (3sin \theta - 4sin {}^{3} \theta) \\ \\ = > \frac{1}{4} sin {}^{3} \theta$

R.H.S

hope it helps ☺️