find sin2x, cos2x, and tan2x if sinx= -2/sqrt 5 and x terminates in quadrant III

Answer 1

if sinx = -2/√5 in III
y = -2, r = √5 , by Pythagoras,
x^2 + 4 = 5
x = -1 in III

cosx = -1/√5
tanx = sinx/cosx = (-2/√)/(-1/√5) = 2

sin 2x = 2sinxcosx = 2(-2/√5)(-1/√5) = 4/5
cos 2x = cos^2 x - sin^2 x = 1/5 - 4/5 = -3/5
tan 2x
= 2tanx/(1 - tan^2 x)
=2(2)/(1 - 4) = -4/3