giving that cosx=4/5 and siny=12/13, find cos(x+y) when x is acute and y is obtuse angle
draw two right - angled triangles
you should recognize the first to be the 3-4-5 and the second as the 5-12-13 Pythagorean triangles
if cosx = 4/5, then sinx = 3/5
if siny = 12/13, then cosy = 5/13
cos(x+y) = cosxcosy - sinx siny
= (4/5)(5/13) - (3/5)(12/13)
= 20/65 - 36/65
= -16/65
Did not notice that y was supposed to be an obtuse angle, that would make y an angle in quadrant II and
cosy = -5/13
make the necessary changes
To find cos(x+y), we will use the trigonometric identity:
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
Given that cos(x) = 4/5 and sin(y) = 12/13, we need to find cos(y) and sin(x) to substitute them into the formula.
To find sin(x), we can use the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
Since we know cos(x) = 4/5, we can substitute it into the equation:
sin^2(x) + (4/5)^2 = 1
Solving this equation, we get:
sin^2(x) = 1 - (16/25) = 9/25
Taking the square root of both sides, we find:
sin(x) = ±3/5
Since x is acute (which means it is in the first or second quadrant), the value of sin(x) should be positive. Therefore, sin(x) = 3/5.
Now, let's find cos(y) using the Pythagorean identity:
sin^2(y) + cos^2(y) = 1
Since we know sin(y) = 12/13, we can substitute it into the equation:
(12/13)^2 + cos^2(y) = 1
Solving this equation, we get:
cos^2(y) = 1 - (144/169) = 25/169
Taking the square root of both sides, we find:
cos(y) = ±5/13
Since y is obtuse (which means it is in the second or third quadrant), the value of cos(y) should be negative. Therefore, cos(y) = -5/13.
Finally, substitute the values of cos(x), sin(x), cos(y), and sin(y) into the formula for cos(x+y):
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
= (4/5)(-5/13) - (3/5)(12/13)
= -20/65 - 36/65
= -56/65
Therefore, cos(x+y) is equal to -56/65.