Asked 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!
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!
Answers
Answered by
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.
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.
Answered by
DeeDee
Oh, I see now! That's exactly what happened!! Thank you, that's perfect!!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.