Question

sin(2/3 π - 2arctg 4/3)

Answer 1

$sin( dfrac{2 pi }{3} -2arctg dfrac{4}{3})$

Пусть $arctg dfrac{4}{3}=x$
$sin( dfrac{2 pi }{3}-2x)=sin dfrac{2 pi }{3}cos2x-cos dfrac{2 pi }{3}sin2x= dfrac{ sqrt{3}cos2x+sin2x }{2}= \ = dfrac{ sqrt{3}-2 sqrt{3}sin^2x+2sinxcosx }{2}= dfrac{ sqrt{3} }{2}- sqrt{3}sin^2x+sinxcosx \ \ tgx= dfrac{4}{3} Rightarrow cosx= sqrt{ dfrac{1}{1+tg^2x} }= sqrt{ dfrac{1}{1+ frac{16}{9} } } = dfrac{3}{5}$
$sinx= sqrt{1-cos^2x}= sqrt{1- dfrac{9}{25} }= dfrac{4}{5} \ \ dfrac{ sqrt{3} }{2}- sqrt{3}sin^2x+sinxcosx= dfrac{ sqrt{3} }{2}- sqrt{3}cdot dfrac{16}{25}+ dfrac{4}{5}cdot dfrac{3}{5} = dfrac{12}{25}- dfrac{7 sqrt{3} }{50}$

Ответ: $dfrac{12}{25}- dfrac{7 sqrt{3} }{50}$