It is known that COS is equal to 12 / 13 of negative, and θ belongs to (π, 3 / 2 π) to find sin θ (θ + 4 π)

It is known that COS is equal to 12 / 13 of negative, and θ belongs to (π, 3 / 2 π) to find sin θ (θ + 4 π)

Cos θ = - 12 / 13 θ belongs to (π, 3 / 2 π), then sin θ = - 5 / 13
sinθ（θ+π/4）=sinθcosπ/4+cosθsinπ/4
=-5/13*√2/2-12/13*√2/2
=-17√2/2 6

3-cos 20 ° / 2-sin ^ 2 80 ° equals

(3-cos 20°)/(2-sin ^2 80°)=[4-(2cos 10°)^2]/[2-(cos10°)^2]=2

Given that sin α = 3 / 5, cos β = - 5 / 13, α belongs to (π / 2), β belongs to (π, 3 π / 2), then cos (α - β) is equal to

I can't figure it out. You can use the square of sin α + the square of cos α = 1 to calculate the values of cos α and sin β, and then use the formula cos (α - β) to expand the calculation