Question

Sendo α e β ângulos do 1°Q, com cos α = 4/5 e cos β = 12/13, calcule:
a) sen (α + β)=

b) cos (α−β)=

c)sen 2β=

d)cos 2α=

Answer 1

Salve Monteirão ( o mais brabo ):

Resolva usando as relações trigonométricas:

sen (a + b) = sen a · cos b + sen b · cos a

sen (a – b) = sen a · cos b – sen b · cos a

cos (a + b) = cos a · cos b – sen a · sen b

cos (a – b) = cos a · cos b + sen a · sen b

sen² + cos² = 1

senα² + cosα² = 1

senα² + (4/5)² = 1

senα² + 16/25 = 1

senα² = 1 - 16/25

senα² = 1 - 16/25

senα = $\sqrt{\frac{9}{25} }$

senα = 3/5

senβ² + cosβ² = 1

senβ² + (12/13)² = 1

senβ² + 144/169 = 1

senβ = $\sqrt{\frac{25}{169} }$

senβ = 5/13

a)

sen (α + β) = senα . cosβ + senβ . cosα

sen (α + β) = 3/5 . 12/13 + 5/13 . 4/5

sen (α + β) = 36/65 + 20/65

sen (α + β) = 56/65

b)

cos (α−β) = cosα . cosβ + senα . senβ

cos (α−β) = 4/5 . 12/13 + 3/5 . 5/13

cos (α−β) = 48/65 + 15/65

cos (α−β) = 63/65

c)

sen 2β = 2 . senβ . cosβ

sen 2β = 2 . 5/13 . 12/13

sen 2β = 2 . 60/169

sen 2β = 120/169

d)

cos 2α = cos² a – sen² a

cos 2α = (4/5)² - (3/5)²

cos 2α = 7/25