Verified answer

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►

► sen(-α) = -sen(α)

► sen(90º+α) = cos(α) , sen(90º-α) = cos(α)

► sen(180º+α) = -sen(α) , sen(180º-α) = sen(α)

► sen(270º+α) = -cos(α) , sen(270º-α) = -cos(α)

► sen(360º+α) = sen(α) , sen(360º-α) = -sen(α)

►

► cos(-α) = cos(α)

► cos(90º+α) = -sen(α) , cos(90º-α) = sen(α)

► cos(180º+α) = -cos(α) , cos(180º-α) = -cos(α)

► cos(270º+α) = sen(α) , cos(270º-α) = -sen(α)

► cos(360º+α) = cos(α) , cos(360º-α) = cos(α)

►

► sen(330º) = sen(360 - 30) = -sen(30) = -1/2

► sen(-450º) = sen(-360-90) = sen(-90) = -sen(90) = -1

► tg(120º) = tg(180-60) = sen(180-60)/cos(180-60)

► tg(120º) = -sen(60)/cos(60) = -√3/2/(1/2) = -√3

► cotg(210º) = cotg(270-60) = cos(270º-60)/sen(270º-60)

► cotg(210º) = -sen(60)/-cos(60) = tg(60) = √3

►

► E = (sen(330°) + sen(-450º))/(tg(120°).cotg(-210°))

► E = (-1/2 -1)/(-√3.√3) = (-3/2)/(-3) = 1/2

►

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sen330=-sen30=-1/2

sen(-450)=-sen90=-1

tg120=sen60/-cos60=(V3/2)/(-1/2)=-V3

ctg(-210)=cos150/sen150=-cos30/sen30=(-V3/2)/1/2)=-V3

sen330°+sen(-450°)/tg120°.ctg(-210)=

=[(-1/2)(-1]/(-V3)(-V3)=-3/2/3=-1/2

sen 330° = - sen 30° = - 1/2

sen (- 450°) = sen (- 90°) = - 1

tg 120° = tg 30° = (raiz de 3) / 3

cot (-210°) = 1 / tg (-210°) = 1 / (- tg 30°) = - 1 / (raiz de 3) / 3 = - 3 / (raiz de 3) = - 3*(raiz de 3) / 3 = - (raiz de 3) / 3.

---Agora basta colocar os valores na fórmula:

(- 1/2 - 1) / [(raiz de 3) / 3]*[- (raiz de 3) / 3]

- 3/2 / [- 3/9]

3/2 / 3/9

9/2.