Question

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

Answer 1

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

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

$3)\quad sin10\cdot sin50\cdot sin70=sin10\cdot sin(60-10)\cdot sin(60+10)=\\\\=Formyla:\star \; \; sin \alpha \cdot sin(60- \alpha )\cdot sin(60+ \alpha )=\frac{1}{4}sin3 \alpha \; \; \; \star =\\\\= \frac{1}{4} sin(3\cdot 10)=\frac{1}{4}sin30=\frac{1}{4}\cdot \frac{1}{2}=\frac{1}{8}\\\\4)\quad cos10\cdot cos50\cdot cos70=cos10\cdot cos(60-10)\cdot cos(60+10)=\\\\=Formyla:\star \; \; cos \alpha \cdot cos(60- \alpha )\cdot sin(60+ \alpha )=\frac{1}{4}cos3 \alpha \; \; \star =$ 

$= \frac{1}{4}cos(3\cdot 10)=\frac{1}{4}cos30=\frac{1}{4} \cdot \frac{\sqrt3}{2}=\frac{\sqrt3}{8}\\\\5)\quad tg40\cdot tg20\cdot tg80=tg20\cdot tg(60-20)\cdot tg(60+20)=\\\\=Formyla:\; \star \; \; tg \alpha \cdot tg(60- \alpha )\cdot tg(60+ \alpha )=tg3 \alpha \; \; \star =\\\\=tg(3\cdot 20)=tg60=\sqrt3$