Posted by Yulieth on .
given sinθ=5/13 and π<θ<3π/2 find
sin2θ
cos( θ4π/3)
sin(θ/2)
can some1 explain to me how to do these?

math 
Henry,
sinA = 5/13 = Y/r,
X^2 + Y^2 = r^2,
X^2 + (5)^2 = (13)^2,
X^2 + 25 = 169,
X^2 = 169  25 = 144,
X = + 12.
X = 12 Because it places our resultant
vector in the required quadrant(Q3).
(X,Y) = (12,5).
tanA = Y/X = 5/12 = 0.416666,
A = 22.62 Deg. ,
A = 22.62 + 180 = 202.62(Q3).
180 < 202.62 < 270. Q3.
sin(2A) = sin(405.24) = 0.710.
cos(A4pi/3) = cos(202.62240) = 0.7946
sin(A/2) = + sqrt((1cosA)/2) =
sqrt((1cos202.62)/2) = 0.9806 
math 
Henry,
Correction:
sinA = 5/13 = Y/r,
A = 157.38 Deg.
sin(2A) = sin(314.76) = 0.710.
cos(A4pi/3)=cos(157.38240)=0.1284
sin(A/2) = + sqrt((1cosA)/2) =
sqrt((1cos157.38)/2 = 0.9806.