Help With Pre-Calculus!! Half/Double Angles and sum/difference trig identities

Sinα = 4/5 and Cosβ = 5/13

Sin2α
Cos2α
Sinαβ
Tan2α
Cos1/2β
Tan1/2α
Sin1/2α
Can You Please show me how to do this practice problems step by step!! :) i have a test on this tomorrow and can seem to understand these!! :/ I would really appreciate it :D

i will assume that we are in the first quadrant.

Sin(α) = 4/5 = y/z

x^2 + y^2 = z^2
x^2 + 4^2 = 5^2
x^2 = 25 - 16
x^2 = 9
x = 3

which implies: ----> cos(α) = 3/5 = x/z

Cos(β) = 5/13 = x/z

x^2 + y^2 = z^2
5^2 + y^2 = 13^2
y^2 = 169 - 25
y^2 = 144
x = 12

which implies: ----> sin(β) = 12/13 = y/z

Let's apply them:

Sin(2α) = 2sin(α)cos(α)
Sin(2α) = 2 * (4/5) * (3/5)
Sin(2α) = (24/25)

cos(2α) = 2cos^2(α) - 1
cos(2α) = 2 * (3/5)^2 - 1
cos(2α) = 2 * (9/25) - 1
cos(2α) = (18/25) - 1
cos(2α) = - (7/25)

Sin(α + β) = sin(α) cos(β) + sin(β) cos(α)
Sin(α + β) = (4/5) (5/13) + (12/13) (3/5)
Sin(α + β) = (4/13) + (36/65)
Sin(α + β) = (56/65)

Sin(α - β) = sin(α) cos(β) - sin(β) cos(α)
Sin(α - β) = (4/5) (5/13) - (12/13) (3/5)
Sin(α - β) = (4/13) - (36/65)
Sin(α - β) = - (16/65)

Tan(2α) = sin(2α) / cos(2α)
Tan(2α) = (24/25) / - (7/25)
Tan(2α) = - (24/7)

Cos(β/2) = +/- √( 1 + cos(β) ) -----> i will assume first quadrant
Cos(β/2) = √( 1 + (5/13) )
Cos(β/2) = √( (18/13) )

sin(α/2) = +/- √( 1 - cos(α) ) -----> i will assume first quadrant
sin(α/2) = √( 1 - (3/5) )
sin(α/2) = √( (2/5) )

Cos(α/2) = +/- √( 1 + cos(α) ) -----> i will assume first quadrant
Cos(α/2) = √( 1 + (3/5) )
Cos(α/2) = √( (8/3) )

tan(α/2) = sin(α/2) / cos(α/2)
tan(α/2) = √( (2/5) ) / √( (8/3) )
tan(α/2) = √( (2/5) / (8/3) )
tan(α/2) = √( (2/5) * (3/8) )
tan(α/2) = √( (3/20) )

=========

free to e-mail if have a question