The first question sinα=4/5, cos(α+β)=5/ 13.
Then cos α = √ [1-(4/5) 2] = 3/5.
sin(α+β)=√[ 1-(5/ 13)^2]= 12/ 13
Then sin β = sin [(α+β)-α] = sin (α+β) cos α-cos (α+β) sin α = (12/13) * (3/5)-(5/13).
The second question sinβ= number of roots1010, so cos β = number of roots10/0.
sin2β=2sinβcosβ=3/5
Sinβ= root number 10/ 10 < 1/2.
So β < 30, so 2β is also an acute angle.
So cos2β=4/5.
tan2β=3/4
tan(α+2β)=(tanα+tan 2β)/( 1-tanα* tan 2β)
=( 1/7+3/4)/[ 1-( 1/7)(3/4)]
=(25/28)/( 1-3/28)
= 1
Question 4 (1) Prove from right to left
Right = (1-tan α tan β) tan (α+β) = (1-tan α tan β) * (tan α+tan β)/(1-tan α tan β) = tan α+tan β = left.
(2)tan 60 = tan(20+40)=(tan 20+tan 40)/( 1-tan 20 tan 40)
tan60=√3
Namely: (tan 20+tan 40)/(1-tan 20 tan 40) = √ 3.
tan 20+tan 40 =√3( 1-tan 20 tan 40)
tan20+tan40=√3-√3tan20tan40
tan20+tan40+√3tan20tan40=√3
(3) The original formula = 1+tanα*tanβ-(tanα+tanβ).
= 2-[(tanα+tanβ)+ 1-tanα* tanβ]
= 2-( 1-tanα* tanβ)( 1+tan(α+β))
=2-0
=2
(4) tan 60 = (tan 20+tan 40)/(1-tan 20 * tan 40) = root number 3.
Tan20+tan40 = root number 3 *( 1-tan20 * tan40)
Substitute into the above formula
(tan 20+tan 40+tan 120) divided by (tan20 tan40)
= (root number 3- root number 3 * tan 20 * tan 40- root number 3)/(tan20 * tan40)
=-root number 3