sin a=3/5, a lies on quadrant 1 and sin b=5/13, b lies on quadrant 2. Find cos(a+b)
To find cos(a+b), we can use the cosine of the sum formula. The formula is as follows:
cos(a+b) = cos a * cos b - sin a * sin b
Let's substitute the given values into the formula.
Since sin a = 3/5 and a lies in quadrant 1, we can use the Pythagorean theorem to find cos a:
cos a = √(1 - sin^2 a) = √(1 - (3/5)^2) = √(1 - 9/25) = √(16/25) = 4/5
Similarly, since sin b = 5/13 and b lies in quadrant 2, we can use the Pythagorean theorem to find cos b:
cos b = √(1 - sin^2 b) = √(1 - (5/13)^2) = √(1 - 25/169) = √(144/169) = 12/13
Now we can substitute the values into the cosine of the sum formula:
cos(a+b) = cos a * cos b - sin a * sin b
= (4/5) * (12/13) - (3/5) * (5/13)
= (48/65) - (15/65)
= 33/65
Therefore, cos(a+b) = 33/65.
To find the value of cos(a+b), we can use the trigonometric identity for the sum of angles:
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
First, let's find the value of cos(a) and sin(b):
Given that sin(a) = 3/5 and a lies in quadrant 1, we can use the Pythagorean identity to find cos(a):
cos(a) = sqrt(1 - sin^2(a))
cos(a) = sqrt(1 - (3/5)^2)
cos(a) = sqrt(1 - 9/25)
cos(a) = sqrt(16/25)
cos(a) = 4/5
Similarly, given that sin(b) = 5/13 and b lies in quadrant 2, we can use the Pythagorean identity to find cos(b):
cos(b) = sqrt(1 - sin^2(b))
cos(b) = sqrt(1 - (5/13)^2)
cos(b) = sqrt(1 - 25/169)
cos(b) = sqrt(144/169)
cos(b) = 12/13
Now we have cos(a) = 4/5 and cos(b) = 12/13. We can substitute these values into the formula to find cos(a+b):
cos(a+b) = (4/5)(12/13) - (3/5)(5/13)
cos(a+b) = 48/65 - 15/65
cos(a+b) = 33/65
Therefore, cos(a+b) = 33/65.
cos(a+b)=sinasinb-cosa*cosb
sina is given, so cosa=4/5
sinb is given so cos b=12/13