If cos A=3/5,cos B =4/5,find the value of cos(A-B)/2
wondering if this is 1/2 *cos(A-B), or cos ((A+B)/2)
probably the latter, as the previous post used half-angle formula.
So, eku, I guess that gives a clue to this one, too, eh?
To find the value of cos(A-B)/2, we need to use the cosine addition or subtraction formula. The formula states that:
cos(A - B) = cos(A) * cos(B) + sin(A) * sin(B)
Let's break down the steps:
1. Determine sin(A):
To find sin(A), we can use the Pythagorean identity: sin^2(A) + cos^2(A) = 1.
Since we know that cos(A) = 3/5, we can substitute it into the equation and solve for sin(A):
sin^2(A) + (3/5)^2 = 1
sin^2(A) + 9/25 = 1
sin^2(A) = 1 - 9/25
sin^2(A) = 16/25
Taking the square root of both sides, we get:
sin(A) = 4/5
2. Determine sin(B):
Similar to the previous step, we can find sin(B) using the Pythagorean identity.
Since cos(B) = 4/5, we have:
sin^2(B) + (4/5)^2 = 1
sin^2(B) + 16/25 = 1
sin^2(B) = 1 - 16/25
sin^2(B) = 9/25
Taking the square root, we get:
sin(B) = 3/5
3. Plug the values into the cosine addition formula:
cos(A - B) = cos(A) * cos(B) + sin(A) * sin(B)
cos(A - B) = (3/5) * (4/5) + (4/5) * (3/5)
cos(A - B) = 12/25 + 12/25
cos(A - B) = 24/25
4. Finally, find cos(A - B)/2:
cos(A - B)/2 = (24/25)/2
cos(A - B)/2 = 24/50
cos(A - B)/2 = 12/25
Therefore, the value of cos(A - B)/2 is 12/25.