Тригонометрични функции

[Есра]

Намерете:
Tg15°
Cos165°

[Добромир Глухаров]

$sin(2\alpha)=\frac{2tg\alpha}{1+tg^2\alpha};\alpha=15^\circ,tg15^\circ=x\Rightarrow\frac{2x}{1+x^2}=sin30^\circ=\frac{1}{2}$

$4x=1+x^2;\ x^2-4x+1=0;\ D_1=2^2-1.1=3;\ x_{1,2}=2\pm\sqrt{3}$

$sin15^\circ<cos15^\circ\Rightarrow tg15^\circ<1\Rightarrow tg15^\circ=x_2=2-\sqrt{3}$

[Добромир Глухаров]

$cos165^\circ=cos(180^\circ-15^\circ)=-cos15^\circ=-\sqrt{\frac{1+cos30^\circ}{2}}=$

$=-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=-\sqrt{\frac{2+\sqrt{3}}{4}}=-\frac{1}{4}\sqrt{8+4\sqrt{3}}=$

$=-\frac{1}{4}\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2.\sqrt{6}.\sqrt{2}}=-\frac{1}{4}\sqrt{(\sqrt{6}+\sqrt{2})^2}=-\frac{\sqrt{6}+\sqrt{2}}{4}$

[ева]

tg 15[tex]^\circ[/tex]=?
(2 начин) tg [tex]\alpha[/tex]=[tex]\frac{1-cos 2\alpha}{sin 2\alpha}[/tex] [tex]\Rightarrow[/tex]
tg15[tex]^\circ[/tex]=[tex]\frac{1-cos 30^\circ}{sin 30^\circ}[/tex]=[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}[/tex]=[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}[/tex]=[tex]\frac{(2-\sqrt{3})2}{1.2}[/tex]=2-[tex]\sqrt{3}[/tex]