Question

математика.. пожалуйста помогите срочно надо ....

Answer 1

$1); ; -3sin^2x+1-3cos^2x=-3(sin^2x+cos^2x)+1=-3+1=-2\\2); ; tg100+frac{sin530}{1+sin640}=tg(90+10)+frac{sin(360+170)}{1+sin(720-80)}=\\=-ctg10+frac{sin(180-10)}{1-sin80}=-ctg10+frac{sin10}{1-sin(90-10)}=\\=-ctg10+frac{sin10}{1-cos10}=frac{sin10}{2sin^220}-ctg10$

$3); ; tga=2\\ frac{3sin^2a+12sinacdot cosa+cos^2a}{sin^2a+sinacdot cosa-2cos^2a}=frac{cos^2acdot (3tg^2a+12tga+1)}{cos^2acdot (tg^2a+tga-2)}=\\=frac{3cdot 2^2+12cdot 2+1}{2^2+2-2}=frac{37}{4}=9,25\\4); ; sin(frac{pi}{2}-alpha )=cosalpha \\cos72^circ = frac{sqrt5-1}{4}$

$star ; ; cos72^circ =cos(90^circ -18^circ )=sin18^circ \\4sin^218=frac{2cdot sin18cdot 2sin18cdot cos18}{cos18}=frac{2cdot sin18cdot sin36}{cos18} =frac{2cdot frac{1}{2}(cos18-cos54)}{cos18}=\\=1-frac{cos54}{cos18}=1-frac{cos(90-36)}{cos18}=1-frac{sin36}{cos18}=1-frac{2sin18cdot cos18}{cos18}=\\=1-2sin18; ; to; ; 4sin^218=1-2sin18\\4sin^218+2sin18-1=0; ,\\t=sin18^circ >0; ,; ; 4t^2+2t-1=0; ,\\D/4=1+4=5; ,; ; ; t=frac{-1pm sqrt5}{4}; ,; ; t>0; ; to \\sin18^circ = frac{-1+sqrt5}{4}=cos72^circ$