given that sinx=3/4 and cosy=-5/13 and both x and y are in quadrant II, find the exact value of cos(x-y)
cos (x-y) = cos x cos y = sin x sin y
cos x = -sqrt7 /4
cos y = -5/13
sin x = 3/4
sin y = 12/13
so
cos(x-y)=5sqrt7/52 - 9/13
sorry typo
cos x cos y + sin x sin y
5 sqrt 7/52 + 9/13
= .9467
sinxeqal to-3/4atquaderant3,cosy equal to7/10at quaderant4 ,then what is:-A,csc/x-y/ B ,cos/x-y/
To find the exact value of cos(x-y), we need to use the difference-of-angles formula for cosine. The formula states that cos(x-y) = cos(x)cos(y) + sin(x)sin(y).
Since we are given sin(x) = 3/4 and cos(y) = -5/13, we need to find cos(x) and sin(y) in order to use the formula.
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can solve for cos(x):
sin^2(x) + cos^2(x) = 1
(3/4)^2 + cos^2(x) = 1
9/16 + cos^2(x) = 1
cos^2(x) = 1 - 9/16
cos^2(x) = 16/16 - 9/16
cos^2(x) = 7/16
Since x is in quadrant II, cos(x) is negative. Therefore, cos(x) = -sqrt(7)/4.
Next, we need to find sin(y). We can use the Pythagorean identity sin^2(y) + cos^2(y) = 1:
sin^2(y) + (-5/13)^2 = 1
sin^2(y) + 25/169 = 1
sin^2(y) = 1 - 25/169
sin^2(y) = 169/169 - 25/169
sin^2(y) = 144/169
sin(y) = sqrt(144)/13
sin(y) = 12/13
Now that we have cos(x), sin(y), and the given values of sin(x) and cos(y), we can substitute them into the difference-of-angles formula:
cos(x-y) = cos(x)cos(y) + sin(x)sin(y)
cos(x-y) = (-sqrt(7)/4)(-5/13) + (3/4)(12/13)
cos(x-y) = sqrt(7)/4 * 5/13 + 3/4 * 12/13
cos(x-y) = 5sqrt(7)/52 + 36/52
cos(x-y) = (5sqrt(7) + 36)/52
Therefore, the exact value of cos(x-y) is (5sqrt(7) + 36)/52.