Question

очень срочно Нужно решение

Answer 1

$frac{tga}{cos^{2}a}-tga=0 \ tga-cos^{2}a*tga=0 \ frac{sina-cos^{2}a*sina}{cosa}=\ sina-cosasina=0 \ sina(1-cosa)=0 \ sina=0 \ a= pi n \ 1-cosa=0 \ cosa=1 \ a= pi +2 pi n \ cos^{2} neq 0 \ cosa neq 0$


$cos225+sin405+tg420= \ cos(180+45)+sin(360+45)+ frac{sin(360+60)}{cos(360+60)}= \ \ -cos45+sin45+ frac{sin60}{cos60}=- frac{ sqrt{2}}{2}+ frac{ sqrt{2}}{2}+ frac{ frac{ sqrt{3}}{2}}{ frac{1}{2}}= frac{ sqrt{3}}{2}*2= sqrt{3}$


$3,5sina-1,5 \ cosa=- frac{4 sqrt{3}}{7} \ frac{ pi }{2}leq a leq pi \ \ sina= sqrt{1-(- frac{4 sqrt{3}}{7} )^{2}}= sqrt{1- frac{48}{49}}= sqrt{ frac{1}{49}}= frac{1}{7} \ 3,5* frac{1}{7}-1,5= frac{7}{2}* frac{1}{7}-1,5=0,5-1,5=-1$


$sin( frac{ pi }{3}-x)+ frac{1}{2}sinx= \ sinfrac{ pi }{3}cosx-cosfrac{ pi }{3}sinx+ frac{1}{2}sinx= \ frac{ sqrt{3}}{2}cosx- frac{1}{2}sinx+frac{1}{2}sinx=frac{ sqrt{3}}{2}cosx$


$sin^{2}( frac{3 pi }{2}+a)-sin^{2}( pi -a) \ \ a= frac{5 pi }{12} \ \ sin^{2}( frac{3 pi }{2}+ frac{5 pi }{12})-sin^{2}( pi - frac{5 pi }{12}) = \ \ sin^{2} frac{23 pi }{12}-sin^{2} frac{7 pi }{12}|*2 \ \ 2sin^{2} frac{23 pi }{12}-2sin^{2} frac{7 pi }{12}= \ \ 1-cos frac{46 pi }{12}-1-cos frac{14 pi }{12}= \ \ -2cos frac{8 pi }{3}=-2*0,5=-1$

$sqrt{6} frac{sin10cos55+sin280sin55}{sin10cos110+sin260cos200}= \ \ sqrt{6} frac{sin10cos55+sin(270+10)sin55}{sin10cos(90+20)+sin(270-10)cos(180+20)}= \ \ sqrt{6} frac{sin10cos55-cos10sin55}{-sin10sin20+cos10cos20}= \ \ sqrt{6} frac{sin10cos55-cos10sin55}{cos10cos20-sin10sin20}= \ \ sqrt{6} frac{sin(10-55)}{cos(10+20)}= \ \ sqrt{6} frac{-sin45}{cos30}=$$\ sqrt{6} frac{- frac{ sqrt{2}}{2}}{ frac{ sqrt{3}}{2}}= \ \ sqrt{6}*- frac{ sqrt{2}}{2}* frac{2}{ sqrt{3}}=- frac{ sqrt{12}}{ sqrt{3}}= -sqrt{4}=-2$


$frac{3(1-2sin^{2}58)}{4sin103*cos77}= \ \ frac{3(1-(1-cos116)}{4sin103cos77}= \ \ frac{3(1-1+cos116)}{4sin(180-77)cos77}= \ \ frac{3cos116}{4sin77cos77}= \ \ frac{3cos116}{2sin154}= \ \ frac{3cos(90+26)}{2sin(180-26)}= frac{-3sin26}{2sin26}=- frac{3}{2}=-1,5$