Tigonometry Help PLEASE?

Question

Find the exact value of cos(x+y), given sin x= 3/5, cos y = 4/5, x is in quadrant II and y is in quadrant I.

I know you do cos(x+y)= cos x cos 4/5 - sin 3/5 sin y

but i dont know what to do from there

please help

Answer 1

You must remember, sear it into your conciousness, that sin and cosine are related by Pythogrean theory Sin^2 x + Cos^2 x = 1. So if sin x= 3/5, cos x=-4/5, given x in quadrant II.
The rest should be yummy duck soup.

Answer 2

OK, you finished the hardest step, the next one is very simple,
all you have to do is:
1. draw a right angled triangle.

2. name any of the two small angles x

3. sin x=3/5, this means the side opposite to x equals 3 in length, and the side opposite to 90 degree angle equal 5 in length, write the numbers on your drawing

4. find the missing side length by Pythagoras rule, it should equal 4 in length.

5. cos x is the side length of 4 over the 90 degree angle side of 5 which makes cos x=4/5

simple, right?

for y, Notice that cos x=4/5 = cos y
this means x=y
which means sin y = sin x

and voila!

but remember, it's just a coincidence that x=y, if not, then u should draw another triangle for y

Answer 3

OK, a bit rusty on my trig, so double check this:

this is a 3-4-5 right triangle,
if you know that sin x= 3/5, then x=(sin^-1)(3/5). x= 36.87 (degrees)
if you know that cos y=4/5, then y=(cos^-1)(4/5). y=36.87
x and y are the same angle

so, since 36.87 + 36.87 = 73.74 (degrees), then:
cos (x+y)= cos (73.74) = .279

again, it's been a long time since I've done trig, so double check that

Answer 4

Unless this is on a test ask you teacher, that is what they get paid for. The teacher should explain all the steps and why.