If sin(A) = 3/5 and cos(B) = 12/13, find sin(A+B) and cos(A-B).

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Step-by-Step Solution

Step 1: Find cos(A) and sin(B)

To find \sin(A+B) and \cos(A-B)$, we need the values of \cos(A) and \sin(B)$. We can find these using the Pythagorean identity \sin^2(\theta) + \cos^2(\theta) = 1$. Assuming angles A and B are acute, \cos(A) and \sin(B) will be positive.

Step 2: Calculate cos(A) and sin(B)

Substitute the given values of \sin(A) and \cos(B) into the Pythagorean identities. Calculate the squares, subtract from 1, and then take the square root to find \cos(A) and \sin(B)$.

Step 3: Apply the sum formula for sin(A+B)

Now we use the sum formula for sine, which states that \sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$. We have all the necessary values to substitute into this formula.

Step 4: Calculate sin(A+B)

Substitute the calculated values of \sin(A)$, \cos(B)$, \cos(A)$, and \sin(B) into the sum formula for sine and perform the multiplication and addition.

Step 5: Apply the difference formula for cos(A-B)

Next, we use the difference formula for cosine, which states that \cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$. We will substitute our known values into this formula.

Step 6: Calculate cos(A-B)

Substitute the calculated values of \cos(A)$, \cos(B)$, \sin(A)$, and \sin(B) into the difference formula for cosine and perform the multiplication and addition.