If sinx=3/5,cosy=-12/13,where x and y both lie in 2nd quad.Find the value os sin(x+y).
luckily, you have nice easy 3-4-5 and 5-12-13 triangles.
cosx = -4/5
siny = 5/13
sin(x+y) = (3/5)(-12/13) + (-4/5)(5/13) = -56/65
reasonable, since (x+y) will be in QIII or QIV
I don't know this answer
To find the value of sin(x+y), we can use the sum of angles formula for sine.
The formula states: sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
Given that sin(x) = 3/5 and cos(y) = -12/13, we need to find the values of cos(x) and sin(y) in order to calculate sin(x + y).
Since x lies in the 2nd quadrant, the value of cos(x) will be negative.
We can use the Pythagorean identity to find cos(x):
cos²(x) = 1 - sin²(x)
cos²(x) = 1 - (3/5)²
cos²(x) = 1 - 9/25
cos²(x) = 25/25 - 9/25
cos²(x) = 16/25
Taking the square root on both sides, we get:
cos(x) = -4/5
Now let's find sin(y):
Using the Pythagorean identity: sin²(y) = 1 - cos²(y)
sin²(y) = 1 - (-12/13)²
sin²(y) = 1 - 144/169
sin²(y) = 169/169 - 144/169
sin²(y) = 25/169
Taking the square root on both sides, we get:
sin(y) = 5/13
Now we can substitute the values in the sum of angles formula:
sin(x + y) = sin(x) * cos(y) + cos(x) * sin(y)
sin(x + y) = (3/5) * (-12/13) + (-4/5) * (5/13)
sin(x + y) = -36/65 - 20/65
sin(x + y) = -56/65
Therefore, the value of sin(x + y) is -56/65.
To find the value of sin(x+y), we can use the following trigonometric identity:
sin(x+y) = sinxcosy + cosxsiny
Since we are given the values of sinx and cosy, we can substitute these into the formula to find sin(x+y).
sin(x+y) = (sinx)(cosy) + (cosx)(siny)
Given sinx = 3/5 and cosy = -12/13, we need to find cosx and siny.
Since x and y lie in the 2nd quadrant, sinx and cosy will both be positive, while cosx and siny will be negative.
We can use the Pythagorean identity to find cosx:
sin^2(x) + cos^2(x) = 1
(3/5)^2 + cos^2(x) = 1
9/25 + cos^2(x) = 1
cos^2(x) = 16/25
cosx = -4/5 (since cosx is negative in the 2nd quadrant)
Similarly, we can use the Pythagorean identity to find siny:
sin^2(y) + cos^2(y) = 1
(12/13)^2 + cos^2(y) = 1
144/169 + cos^2(y) = 1
cos^2(y) = 25/169
cosy = -5/13 (since cosy is negative in the 2nd quadrant)
Now, substitute the values of sinx, cosy, cosx, and siny into the formula:
sin(x+y) = (3/5)(-12/13) + (-4/5)(-5/13)
sin(x+y) = -36/65 + 20/65
sin(x+y) = -16/65
Therefore, the value of sin(x+y) is -16/65.