Вычислите а) √2 tg(240)*sin(300)*cos(1305) б) cos(225)*sin(300)+sin^3(45)*ctg(150) в) cos(1110)*sin(1125) г) sin(25)*sin(85)+sin(65)*sin(5) д) (sin(26)*sin(20)+cos(26)*cos(20))/2cos(6) е) cos(66)*cos(356)+sin(186)*cos(156) /sin(36)*cos(6)-sin(126)*cos(84) ж) 3cosa-4sina/4sina+5cosa . если tga=-0.625

а) √2 *tq240°*sin300°*cos1305°=
√2 *tq(180°+60°)*sin(360°-60°)*cos(7*180 °+45°)=√2 *tq60°*(-sin60°)*(-cos45°) =√2 *√3*(-(√3)/2)*(-√2)/2 =3/2 =1,5.

б) cos225°*sin300°+sin³45°*ctq150° =
cos(180°+45°)*sin(360° -60°)+sin³45°*ctq(180°-30°) =
-cos45°*(-sin60°+sin³45°*(-ctq30°) = (√2/2)*(√3/2)-(√2/2)³*√3 =
(√6)/4 - (√6)/4= 0.

в)  cos1110°*sin1125° =cos(3*360°+30°)°*sin(3*360°+45°) =
cos30°°*sin45° =√3/2 *√2/2 = (√6)/4.

г)  sin25°*sin85° +sin65°*sin5° =sin(90°-65°)*sin(90°-5°) +sin65°*sin5° =
cos65°cos5° +sin65°*sin5°=cos(65°-5°)=cos(60°) =0,5.

д)  (sin26°*sin20° + cos26°*cos20°)/2cos6°=
 (cos26°*cos20°+sin26°*sin20° )/2cos6 =
cos(26°-20°)/2cos6°= cos6°/2cos6° =0,5.

е) !  cos356° заменяю  на  cos354° ...
 (cos66°*cos354°+sin186°*cos156°)/(sin36°*cos6° -sin126°*cos84°)=
(cos66°*cos6°+sin6°*sin66°)/(sin36°*cos6° -cos36°*sin6°) =
cos(66°-6°)/sin(36°-6°) =cos60°/sin30° =(1/2) / (1/2) =1.
ж) (3cosα -4sinα)/(4sinα +5cosα)=  || cosα(3 -4tqα)/cosα(4tqα +5)||
(3 -4tqα)/(4tqα +5)  =(3 -4*(-0,625))/(4*(-0,625)+5) =(3+2,5)(-2,5 +5) =
5,5/2,5 =11/5 =2,2.
* * * слишком  много был * * *
От дед мороза !