IIT JEE $8{\sin ^3}\frac{\pi }{{14}} – 4{\sin ^2}\frac{\pi }{{14}} – 4\sin \frac{\pi }{{14}} = $ December 20, 2022 Engineer Manish Verma Solution [.pdf] Share this:Click to share on Twitter (Opens in new window)Click to share on Facebook (Opens in new window) Related
