Trigonometry
posted by DeeDee .
If
Hello, I have been working on this one problem for a while now and my professor is on vacation. Could someone give me a point in the right direction?
Question: If sin x = 2/3 and sec y = 5/3, where x and y lie between 0 and π/2, evaluate sin(x + y).
My answer was: (3+4sqrt5)/15
I got this based off of sinx=2/3, cosx=sqrt5/3, siny=4/5, cosy=3/5.
Why would that not be correct?
Thank you so much for any help!

Trigonometry 
Reiny
I did not get this
sinx = 2/3, then cosx = √5/3
if secy = 5/3, then
cosy = 3/5, and siny = 4/5
sin(x+y) = sinxcosy + cosxsiny
= (2/3)(3/5) + (√5/3)(4/5)
= (6 + 4√5)/15
Your individual trig values are correct, you must have made a substitution error. 
Trigonometry 
DeeDee
Oh, I see now! That's exactly what happened!! Thank you, that's perfect!!
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