Question

cos п/12
tg п/12
tg п/8
​

Answer 1

Ответ:

$1)\ \ cos\dfrac{\pi}{12}=cos15^\circ =cos(45^\circ -30^\circ )=cos45^\circ \cdot cos30^\circ +sin45^\circ \cdot sin30^\circ =\\\\\\=\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt3+1}{2\sqrt2}=\dfrac{\sqrt6+\sqrt2}{4}$

$2)\ \ tg\dfrac{\pi}{12}=tg15^\circ =tg(45^\circ -30^\circ )=\dfrac{tg45^\circ -tg30^\circ }{1+tg45^\circ \cdot tg30^\circ }=\dfrac{1-\dfrac{\sqrt3}{3}}{1+\dfrac{\sqrt3}{3}}=\\\\\\=\dfrac{3-\sqrt3}{3+\sqrt3}}=\dfrac{\sqrt3-1}{\sqrt3+1}$

$3)\ \ tg\dfrac{\pi}{8}=\dfrac{sin\dfrac{\pi}{8}}{cos\dfrac{\pi}{8}}=\dfrac{\sqrt{\dfrac{1-cos\frac{\pi}{4}}{2}}}{\sqrt{\dfrac{1+cos\dfrac{\pi}{4}}{2}}}=\dfrac{\sqrt{1-\dfrac{\sqrt2}{2}}}{\sqrt{1+\dfrac{\sqrt2}{2}}}=\sqrt{\dfrac{2-\sqrt2}{2+\sqrt2}}=\sqrt{\dfrac{\sqrt2-1}{\sqrt2+1}}\\\\\\\star \ \ \ sin^2x=\dfrac{1-cos2x}{2}\ \ ,\ \ cos^2x=\dfrac{1+cos2x}{2}\ \ \star$