Тест по теме: Тангенс суммы и разности аргументов

\text{tg}(x+y)=\cfrac{\text{tg} x+\text{tg} y}{1-\text{tg} x \text{tg} y}

\text{tg}(x-y)=\cfrac{\text{tg} x-\text{tg} y}{1+\text{tg} x \text{tg} y}

Найдите значение \mathrm{tg~}75^{\circ}

\mathrm{tg~}75^{\circ}=

\mathrm{tg}\left(45^{\circ}+30^{\circ}\right)=

\mathrm{\cfrac{tg\;45^{\circ}+\mathrm{tg}30^{\circ}}{1-\mathrm{tg}\mathrm{~45^{\circ}}\operatorname{tg}30^{\circ}}}=

\cfrac{1+\cfrac{\sqrt{3}}{3}}{1-1\cdot\cfrac{\sqrt{3}}{3}}\,=

\cfrac{3+\sqrt{3}}{3-\sqrt{3}}

\mathrm{tg}\left({\cfrac{\pi}{4}}-\,x\right)=

\cfrac{{\mathrm{tg}}\cfrac{\pi}{4}-{\mathrm{tg}}x}{1+{\mathrm{tg}}\cfrac{\pi}{4}{\mathrm{tg}}x}=

\cfrac{1-\operatorname{tg}\,x}{1+\operatorname{tg}x}

\operatorname{tg} 105^{\circ}= \operatorname{tg}\Big(45^{\circ}+60^{\circ}\Big)=

\cfrac{\operatorname{tg} 45^{\circ}+ \operatorname{tg} 60^{\circ} }{1- \operatorname{tg} 45^{\circ} \cdot \operatorname{tg} 60^{\circ}}=

\cfrac{1+ \sqrt{3} }{1-1 \cdot \sqrt{3}}=

=\cfrac{(1+\sqrt{3})^{2}}{(1-\sqrt{3})( 1+\sqrt{3} )} =

\cfrac{1+2 \sqrt{3}+3}{1-3} =\cfrac{4+2 \sqrt{3}}{-2}=

-2-\sqrt{3} \text {; }

\operatorname{tg} \cfrac{5 \pi}{12}= \operatorname{tg}\Big(\cfrac{3\pi}{12}+\cfrac{2\pi}{12}\Big)=

\operatorname{tg}\Big(\cfrac{\pi}{4}+\cfrac{\pi}{6}\Big)= \cfrac{\operatorname{tg} \cfrac{\pi}{4}+\operatorname{tg} \cfrac{\pi}{6}}{1-\operatorname{tg} \cfrac{\pi}{4} \cdot \operatorname{tg} \cfrac{\pi}{6}}=

\cfrac{1+ \cfrac{\sqrt{3}}{3} }{1-1 \cdot \cfrac{\sqrt{3}}{3}}= \cfrac{3+ \sqrt{3} }{3-\sqrt{3}}=

\cfrac{(3+\sqrt{3})^{2}}{(3-\sqrt{3})(3+\sqrt{3})}=

\cfrac{9+6 \sqrt{3}+3}{9-3}=\cfrac{12+6 \sqrt{3}}{6}=2+\sqrt{3} ;

\operatorname{tg} 165^{\circ}= \operatorname{tg}\Big(45^{\circ}+120^{\circ}\Big)=

\cfrac{\operatorname{tg} 45^{\circ}+\operatorname{tg} 120^{\circ}}{1-\operatorname{tg} 45^{\circ} \cdot \operatorname{tg} 120^{\circ}}=

\cfrac{1+(-\sqrt{3})}{1-1 \cdot(-\sqrt{3})}=

=\cfrac{1-\sqrt{3}}{1+\sqrt{3}}=\cfrac{(1-\sqrt{3})^{2}}{(1+\sqrt{3})(1-\sqrt{3})}=

\cfrac{1-2 \sqrt{3}+3}{1-3}=\cfrac{4-2 \sqrt{3}}{-2}=\sqrt{3}-2 \text {; }

\cfrac{\operatorname{tg} x+\operatorname{tg} 3 x}{1-\operatorname{tg} x \cdot \operatorname{tg} 3 x}=1 ; \\ \operatorname{tg}(x+3 x)=1 ; \\ \operatorname{tg} 4 x=1 ; \\ 4 x=\operatorname{arctg} 1+\pi n=\cfrac{\pi}{4}+\pi n ; \\ x=\cfrac{1}{4} \cdot\Big(\cfrac{\pi}{4}+\pi n\Big)=\cfrac{\pi}{16}+\cfrac{\pi n}{4} ; \\ \text { Ответ: } \cfrac{\pi}{16}+\cfrac{\pi n}{4} . \\

\cfrac{\operatorname{tg} 5 x-\operatorname{tg} 3 x}{1+\operatorname{tg} 3 x \cdot \operatorname{tg} 5 x}=\sqrt{3} ; \\ \operatorname{tg}(5 x-3 x)=\sqrt{3} ; \\ \operatorname{tg} 2 x=\sqrt{3} ; \\ 2 x=\operatorname{arctg} \sqrt{3}+\pi n=\cfrac{\pi}{3}+\pi n ; \\ x=\cfrac{1}{2} \cdot\Big(\cfrac{\pi}{3}+\cfrac{\pi n}{3}\Big)=\cfrac{\pi}{6}+\cfrac{\pi n}{2} ; \\ \text { Ответ: } \cfrac{\pi}{6}+\cfrac{\pi n}{2} .