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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!
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.
Oh, I see now! That's exactly what happened!! Thank you, that's perfect!!
To evaluate sin(x + y), we can use the sum of angles formula for sine:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
Let's go step by step to find the correct answer:
1. Given that sin x = 2/3, we can find cos x using the Pythagorean identity: sin^2(x) + cos^2(x) = 1. Plugging in the value of sin x, we get:
(2/3)^2 + cos^2(x) = 1
4/9 + cos^2(x) = 1
cos^2(x) = 1 - 4/9
cos^2(x) = 5/9
Since x lies between 0 and π/2, cos x must be positive, so we take the positive square root:
cos x = √(5/9) = √5/3
2. Given that sec y = 5/3, we know that sec y is the reciprocal of cos y:
sec y = 1/cos y
So, we have:
5/3 = 1/cos y
Solving for cos y:
cos y = 3/5
Since y lies between 0 and π/2, cos y must be positive.
3. Now that we have sin x, cos x, sin y, and cos y, we can calculate sin(x + y) using the sum of angles formula:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
= (2/3)(3/5) + (√5/3)(4/5)
= 6/15 + 4√5/15
= (6 + 4√5)/15
So, the correct answer is (6 + 4√5)/15.
Comparing this with your answer, we can see that (3 + 4√5)/15 is not correct. There might have been an error in calculating the values of cos x and cos y.
I hope this explanation helps clarify the steps required to solve the problem. Good luck!