Question

Sin 390 cos 420 tg 780 ctg 405

Answer 1

$\sin390\textdegree\cdot \cos420\textdegree\cdot \text{tg}780\textdegree\cdot \text{ctg}405\textdegree=\\\\=\sin(360\textdegree+30\textdegree)\cdot \cos(360\textdegree+60\textdegree)\cdot \text{tg}(4\cdot 180\textdegree+60\textdegree)\cdot \text{ctg}(2\cdot 180\textdegree+45\textdegree)=\\\\=\sin30\textdegree\cdot \cos60\textdegree\cdot \text{tg}60\textdegree\cdot \text{ctg}45\textdegree=\\\\=\dfrac 12\cdot \dfrac 12\cdot \sqrt 3\cdot 1\boldsymbol{=\dfrac{\sqrt3}4}$

====================================

Периодичность тригонометрических функций :

sin (360° + α) = sin α

cos (360° + α) = cos α

tg (180° + α) = tg α

ctg (180° + α) = ctg α