Question

a) cos345° =
b) sin 285° =
c) tg375° =​

Answer 1

Основные формулы:

$\cos(-\alpha )=\cos \alpha$

$\cos(360^\circ+\alpha )=\cos \alpha$

$\sin(270^\circ+\alpha )=-\cos \alpha$

$\mathrm{tg}\, (180^\circ+\alpha )=\mathrm{tg}\, \alpha$

$\cos(\alpha -\beta )=\cos\alpha \cos\beta +\sin\alpha \sin\beta$

$\mathrm{tg}\, (\alpha -\beta )=\dfrac{\mathrm{tg}\, \alpha -\mathrm{tg}\, \beta }{1+\mathrm{tg}\, \alpha\, \mathrm{tg}\, \beta }$

a)

$\cos345^\circ=\cos(360^\circ-15^\circ)=\cos(-15^\circ)=\cos15^\circ=\cos(45^\circ-30^\circ)=$

$=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\dfrac{\sqrt{2} }{2} \cdot\dfrac{\sqrt{3} }{2}+\dfrac{\sqrt{2} }{2} \cdot\dfrac{1 }{2}=\boxed{\dfrac{\sqrt{6}+\sqrt{2} }{4} }$

b)

$\sin285^\circ=\sin(270^\circ+15^\circ)=-\cos15^\circ=\boxed{-\dfrac{\sqrt{6}+\sqrt{2} }{4} }$

c)

$\mathrm{tg}\, 375^\circ=\mathrm{tg}\, (2\cdot180^\circ+15^\circ)=\mathrm{tg}\, 15^\circ=\mathrm{tg}\, (45^\circ-30^\circ)=$

$=\dfrac{\mathrm{tg}\, 45^\circ-\mathrm{tg}\, 30^\circ}{1+\mathrm{tg}\, 45^\circ\mathrm{tg}\, 30^\circ}=\dfrac{1-\dfrac{\sqrt{3} }{3} }{1+1\cdot\dfrac{\sqrt{3} }{3} }=\dfrac{3-\sqrt{3} }{3+\sqrt{3} }=\dfrac{(3-\sqrt{3})^2 }{(3+\sqrt{3})(3-\sqrt{3}) }=$

$=\dfrac{3^2-2\cdot3\cdot\sqrt{3}+(\sqrt{3})^2}{3^2-(\sqrt{3})^2 }=\dfrac{9-6\sqrt{3}+3}{9-3 }=\dfrac{12-6\sqrt{3}}{6 }=\boxed{2-\sqrt{3}}$