suppose sin(x)=4/5 and cos(y)=-12/13, and x and y are in the second quadrant. determine the value of each expression: sin(x+y), cos(x-y), cos2x, sin2y.

To solve this problem, we will use the following trigonometric identities:

1. sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
2. cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y)
3. cos(2x) = cos^2(x) - sin^2(x)
4. sin(2y) = 2 * sin(y) * cos(y)

Given that sin(x) = 4/5 and cos(y) = -12/13, we will find sin(y) and cos(x) using the trigonometric identity for the second quadrant:

1. sin^2(x) + cos^2(x) = 1
(4/5)^2 + cos^2(x) = 1
16/25 + cos^2(x) = 1
cos^2(x) = 1 - 16/25
cos^2(x) = 9/25
cos(x) = ± sqrt(9/25)
cos(x) = ± 3/5

Since x is in the second quadrant, cos(x) is negative:

cos(x) = -3/5

Now, let's find sin(y):

sin^2(y) + cos^2(y) = 1
sin^2(y) + (-12/13)^2 = 1
sin^2(y) + 144/169 = 1
sin^2(y) = 1 - 144/169
sin^2(y) = 25/169
sin(y) = ± sqrt(25/169)
sin(y) = ± 5/13

Since y is in the second quadrant, sin(y) is positive:

sin(y) = 5/13

Now we can substitute the values of sin(x), cos(y), sin(y), and cos(x) into the trigonometric identities to find the values of the expressions:

1. sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
= (4/5) * (-12/13) + (-3/5) * (5/13)
= -48/65 - 15/65
= -63/65

2. cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y)
= (-3/5) * (-12/13) + (4/5) * (5/13)
= 36/65 + 20/65
= 56/65

3. cos(2x) = cos^2(x) - sin^2(x)
= (-3/5)^2 - (4/5)^2
= 9/25 - 16/25
= -7/25

4. sin(2y) = 2 * sin(y) * cos(y)
= 2 * (5/13) * (-12/13)
= -120/169

Therefore, the values of the expressions are:

- sin(x + y) = -63/65
- cos(x - y) = 56/65
- cos(2x) = -7/25
- sin(2y) = -120/169

To determine the value of each expression, we need to use the trigonometric identities and the given information. Let's break down each expression step by step.

1. sin(x+y):
Since sin(x) = 4/5 and x is in the second quadrant, we know that sin(x) is positive. Now, we need to find sin(y) to determine sin(x+y).
To find sin(y), we can use the Pythagorean identity: sin^2(y) + cos^2(y) = 1.
Since cos(y) = -12/13, we can solve for sin(y):
sin^2(y) + (-12/13)^2 = 1
sin^2(y) = 1 - (-12/13)^2
sin^2(y) = 1 - 144/169
sin^2(y) = (169 - 144)/169
sin^2(y) = 25/169
sin(y) = ±√(25/169)
Since y is in the second quadrant (where sin is positive), sin(y) = √(25/169) = 5/13.

Now, we have sin(x) = 4/5 and sin(y) = 5/13.
Using the sum formula for sine, sin(x+y) = sin(x)cos(y) + cos(x)sin(y):
sin(x+y) = (4/5) * (-12/13) + √(25/169) * √(169/169)
sin(x+y) = (-48/65) + (5/13)
sin(x+y) = (-48/65) + (25/65)
sin(x+y) = -23/65

Therefore, sin(x+y) = -23/65.

2. cos(x-y):
Using the difference formula for cosine, cos(x-y) = cos(x)cos(y) + sin(x)sin(y):
cos(x-y) = √(1 - sin^2(x)) * √(1 - sin^2(y)) + (4/5) * (5/13)
cos(x-y) = √(1 - (4/5)^2) * √(1 - (5/13)^2) + (20/65)
cos(x-y) = √(1 - 16/25) * √(1 - 25/169) + (20/65)
cos(x-y) = √(9/25) * √(144/169) + (20/65)
cos(x-y) = (3/5) * (12/13) + (20/65)
cos(x-y) = (36/65) + (20/65)
cos(x-y) = 56/65

Therefore, cos(x-y) = 56/65.

3. cos(2x):
To find cos(2x), we can use the double-angle formula for cosine: cos(2x) = cos^2(x) - sin^2(x).
From the given information, we already know that sin(x) = 4/5.
To find cos(x), we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1.
Substituting sin(x) = 4/5, we can solve for cos(x):
(4/5)^2 + cos^2(x) = 1
16/25 + cos^2(x) = 1
cos^2(x) = 1 - 16/25
cos^2(x) = (25 - 16)/25
cos^2(x) = 9/25
cos(x) = ±√(9/25)
Since x is in the second quadrant (where cos is negative), cos(x) = -√(9/25) = -3/5.

Now, we have sin(x) = 4/5 and cos(x) = -3/5.
Using the double-angle formula for cosine, cos(2x) = cos^2(x) - sin^2(x):
cos(2x) = (-3/5)^2 - (4/5)^2
cos(2x) = 9/25 - 16/25
cos(2x) = -7/25

Therefore, cos(2x) = -7/25.

4. sin(2y):
To find sin(2y), we can use the double-angle formula for sine: sin(2y) = 2sin(y)cos(y).
From the given information, we already know that sin(y) = 5/13 and cos(y) = -12/13.
Using the double-angle formula for sine, sin(2y) = 2(5/13)(-12/13):
sin(2y) = -120/169

Therefore, sin(2y) = -120/169.

To summarize:
- sin(x+y) = -23/65
- cos(x-y) = 56/65
- cos(2x) = -7/25
- sin(2y) = -120/169

you should recognize the 3,4,5 and the 5,12, 13 right-angled triangles, so

if sinx = 4/5, then cosx = -3/5 in II
if cosy = -1213, then siny = 5/13 in II

to have this type of question, you must also have come across the expansions for sin(A+B), cos(A-B), etc.

I will do cos(x-y)

cos(x-y) = cosx cosy + sinx siny
= (-3/5)(-12/13) + (4/5)(5/13)
= 56/65