Đáp án + Giải thích các bước giải:

Ta có: $-555^\circ + 2 \cdot 360^\circ = 165^\circ$ 

$\Rightarrow \begin {cases} \sin (-555^\circ) = \sin 165^\circ \\ \cos (-555^\circ) =\cos 165^\circ \\ \tan (-555^\circ) = \tan 165^\circ \\ \cot (-555^\circ) = \cot 165^\circ \end {cases}$

Mà $165^\circ = 180^\circ - 15^\circ$

$\Rightarrow \begin {cases} \sin (165^\circ) = \sin 15^\circ \\ \cos (165^\circ) = -\cos 15^\circ \\ \tan (165^\circ) = -\tan 15^\circ \\ \cot (165^\circ) = -\cot 15^\circ \end {cases}$

Ta có: $\sin (15^\circ)= \sin (45^\circ - 30^\circ)$

$= \sin 45^\circ \cos 30^\circ - \sin 30^\circ \cos 45^\circ$

$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2}$

$= \dfrac{\sqrt{6}- \sqrt{2}}{4}$

$\Rightarrow \sin^2 (15^\circ) = \dfrac{(\sqrt{6} - \sqrt{2})^2}{16} = \dfrac{2 - \sqrt{3}}{4}$

$\Rightarrow \cos (15^\circ) = \sqrt{1 - \sin^2 (15^\circ)} = \sqrt{1 - \dfrac{2 - \sqrt{3}}{4}} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$

$\Rightarrow \tan (15^\circ) = \dfrac{\sin 15^\circ}{\cos 15^\circ}= \dfrac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} = 2 - \sqrt{3}$

$\Rightarrow \cot (15^\circ) = \dfrac{1}{\tan 15^\circ} = \dfrac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$

$\Rightarrow \begin {cases} \sin (-555^\circ) = \sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4} \\ \cos (-555^\circ) = -\cos 15^\circ = \dfrac{-\sqrt{6} - \sqrt{2}}{4} \\ \tan (-555^\circ) = -\tan 15^\circ = \sqrt{3} - 2 \\ \cot (-555^\circ) = -\cot 15^\circ = -\sqrt{3} - 2 \end {cases}$