Question

3)Relacione:
(a) cos5240° ( )1/2
(b)sen1200°( )-cos 20°
(c)sen(-210°)cos 30°
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Answer 1

$\mathsf{5240^{\circ}=14\times360^{\circ}+200^{\circ}}$

$\mathsf{cos(5240^{\circ})=cos(200^{\circ})=-cos(20^{\circ})}$

$\mathsf{1200^{\circ}=3\times360^{\circ}+120^{\circ}}\\\mathsf{sen(1200^{\circ})=sen(120^{\circ})=sen(60^{\circ})=cos(30^{\circ})}$

$\mathsf{sen(-210^{\circ})=-sen(210^{\circ})=-[-sen(30^{\circ})]=sen(30^{\circ})=\dfrac{1}{2} }$