Question

given sinθ=-5/13 and π<θ<3π/2 find

sin2θ
cos( θ-4π/3)
sin(θ/2)

can some1 please help me with this?

Answer 1

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 half-angle and double-angle and sum/difference formulas.

sin 2θ = 2 sinθ cosθ = 2(-5/13)(-12/13) = 120/169

cos(θ-4π/3) = cosθ cos4π/3 + sinθ sin4π/3
= (-12/13)(-1/2) + (-5/13)(-√3/2)
= 12/26 + 5√3/26

sin θ/2 = √((1-cosθ)/2)
= √((1 + 12/13)/2)
= √(25/26)


Use this same method for all your other similar problems.