Find cos(a=b). sin a = 3/5, a lies in Quadrant II, and cos b = 5/13, b lies in Quadrant I.
since the = and + sign are on the same key, I will assume cos(a=b) is a typo and you meant
cos(a+b)
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
since a is in II and sin(a) = 3/5, cos(a) = -4/5
since b in in I and cos(b) = 5/13, sin(b) = 12/13
so cos(a+b) = (4/5)(5/13) - (3/5)(12/13)
= (-20-36)/65
= -56/65
To find cos(a = b), we need to find the value of cos of the angle that is formed when a and b are equal.
Given that sin(a) = 3/5 and a lies in Quadrant II, we can use the Pythagorean identity sin^2(a) + cos^2(a) = 1 to find cos(a).
First, let's find cos(a):
Since sin(a) = 3/5, and sin^2(a) + cos^2(a) = 1, we have:
(3/5)^2 + cos^2(a) = 1
Simplifying this equation:
(9/25) + cos^2(a) = 1
cos^2(a) = 1 − 9/25
cos^2(a) = 25/25 − 9/25
cos^2(a) = 16/25
Taking the square root of both sides, we get:
cos(a) = ±√(16/25)
Now, since a lies in Quadrant II, where cos(a) is negative, we can use the negative sign:
cos(a) = -√(16/25)
cos(a) = -4/5
Next, let's find cos(b):
Given that cos(b) = 5/13 and b lies in Quadrant I, we already have the value of cos(b).
Finally, to find cos(a = b), we can simply substitute the value of cos(a) and cos(b) to the equation:
cos(a = b) = cos(a) * cos(b)
cos(a = b) = (-4/5) * (5/13)
Simplifying the expression:
cos(a = b) = -20/65
Therefore, cos(a = b) = -4/13.