Posted by Yuli on .
given sinθ=5/13 and π<θ<3π/2 find
sin2θ
cos( θ4π/3)
sin(θ/2)
can some1 please help me with this?

mathematics 
Steve,
Let's go through this again.
θ is in the 3rd quadrant, so x and y are both negative, and the hypotenuse h = 13
sinθ = 5/13
cosθ = 12/13
Use these values in your halfangle and doubleangle and sum/difference formulas.
sin 2θ = 2 sinθ cosθ = 2(5/13)(12/13) = 120/169
cos(θ4π/3) = cosθ cos4π/3 + sinθ sin4π/3
= (12/13)(1/2) + (5/13)(√3/2)
= 12/26 + 5√3/26
sin θ/2 = √((1cosθ)/2)
= √((1 + 12/13)/2)
= √(25/26)
Use this same method for all your other similar problems.