Trigonometry
posted by Elena
find sin2x, cos2x, and tan2x if sinx= 2/sqrt 5 and x terminates in quadrant III

Reiny
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
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