Find cos(a+b). sin a = 3/5, a lies in Quadrant I, and cos b = 12/13, b lies in Quadrant I.
well, we know that
cos a = 4/5
sin b = 5/13
Now just apply the formula
cos(a+b) = cosa cosb - sina sinb
To find cos(a+b), we can use the trigonometric identity:
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
Given that sin a = 3/5 and cos b = 12/13, we can substitute these values into the formula:
cos(a+b) = (cos(a))(cos(b)) - (sin(a))(sin(b))
Let's calculate each term step-by-step:
cos(a) = √(1 - sin²(a)) [using the Pythagorean Identity]
= √(1 - (3/5)²)
= √(1 - 9/25)
= √(16/25)
= 4/5
cos(a) = 4/5
cos(b) = 12/13
Now, let's substitute these values back into the formula:
cos(a+b) = (4/5)(12/13) - (3/5)(sin(b))
cos(a+b) = (48/65) - (3/5)(sin(b))
However, we don't have the value of sin(b) given, so we cannot calculate the exact value of cos(a+b) without that information.
To find cos(a+b), we can use the trigonometric identity for the cosine of the sum of two angles:
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
Given that sin a = 3/5 and cos b = 12/13, we need to find cos a and sin b to substitute them into the formula.
Since a lies in Quadrant I and sin a = 3/5, we can use the Pythagorean identity to find cos a:
cos^2(a) = 1 - sin^2(a)
cos^2(a) = 1 - (3/5)^2
cos^2(a) = 1 - 9/25
cos^2(a) = 16/25
Since a lies in Quadrant I, cos a is positive, so we take the positive square root:
cos(a) = √(16/25)
cos(a) = 4/5
Similarly, since b lies in Quadrant I and cos b = 12/13, we can use the Pythagorean identity to find sin b:
sin^2(b) = 1 - cos^2(b)
sin^2(b) = 1 - (12/13)^2
sin^2(b) = 1 - 144/169
sin^2(b) = 25/169
Since b lies in Quadrant I, sin b is positive, so we take the positive square root:
sin(b) = √(25/169)
sin(b) = 5/13
Now we can substitute the values back into the formula for cos(a+b):
cos(a+b) = (4/5)(12/13) - (3/5)(5/13)
cos(a+b) = 48/65 - 15/65
cos(a+b) = (48 - 15)/65
cos(a+b) = 33/65
Therefore, cos(a+b) = 33/65.