Question

(Срочно!!!)Результат вычисления:
1) 1+ctg15°/1-ctg15°=
2) cos 36°- sin18°=
3) sin10°•sin50°•sin70°=
4) cos10°•cos50°•cos70°=
5) tg40°•tg20°•tg80°=

Answer 1

$1)quad frac{1+ctg15}{1-ctg15} = frac{ctg45+ctg15}{1-ctg45cdot ctg15} =- frac{ctg45+ctg15}{ctg45cdot ctg15-1} =\\=-( frac{ctg45cdot ctg15-1}{ctg45+ctg15} )^{-1}=-(ctg(45+15))^{-1}=-(ctg60)^{-1}=\\=-(frac{1}{sqrt3})^{-1}=-sqrt3\\2)quad sin18^circ= frac{sqrt5-1}{4} \\cos36^circ -sin18^circ =cos(2cdot 18^circ )-sin18^circ =\\=1-2sin^218^circ -sin18^circ =1-2cdot (frac{sqrt5-1}{4} )^2 -frac{sqrt5-1}{4}=$

 $=1- frac{6-2sqrt5}{8} - frac{sqrt5-1}{4} = frac{8-6+2sqrt5-2sqrt5+2}{8} =$  $frac{4}{8}= frac{1}{2}$

$3)quad sin10cdot sin50cdot sin70=sin10cdot sin(60-10)cdot sin(60+10)=\\=Formyla:star ; ; sin alpha cdot sin(60- alpha )cdot sin(60+ alpha )=frac{1}{4}sin3 alpha ; ; ; star =\\= frac{1}{4} sin(3cdot 10)=frac{1}{4}sin30=frac{1}{4}cdot frac{1}{2}=frac{1}{8}\\4)quad cos10cdot cos50cdot cos70=cos10cdot cos(60-10)cdot cos(60+10)=\\=Formyla:star ; ; cos alpha cdot cos(60- alpha )cdot sin(60+ alpha )=frac{1}{4}cos3 alpha ; ; star =$ 

$= frac{1}{4}cos(3cdot 10)=frac{1}{4}cos30=frac{1}{4} cdot frac{sqrt3}{2}=frac{sqrt3}{8}\\5)quad tg40cdot tg20cdot tg80=tg20cdot tg(60-20)cdot tg(60+20)=\\=Formyla:; star ; ; tg alpha cdot tg(60- alpha )cdot tg(60+ alpha )=tg3 alpha ; ; star =\\=tg(3cdot 20)=tg60=sqrt3$