if sinA +sinB=a AND cosA+ cosB=b then prove that sin(A+B)=(2ab)/(a²+b²)

To prove that \( \sin(A+B) = \frac{2ab}{a^2 + b^2} \) given that \( \sin A + \sin B = a \) and \( \cos A + \cos B = b \), we can start with some trigonometric identities and manipulations. Let's break this down step by step.

Step 1: Use the Sine and Cosine Addition Formulas

Recall the sine addition formula, which states:

\( \sin(A+B) = \sin A \cos B + \cos A \sin B \)

We need to express \( \sin A \) and \( \sin B \) in terms of \( a \) and \( b \). Using the identities for sine and cosine, we can derive \( \sin^2 A + \cos^2 A = 1 \) and similarly for \( B \).

Step 2: Express Sine and Cosine in Terms of a and b

From the given equations, we can manipulate them to express \( \sin A \) and \( \sin B \) in a useful way:

• From \( \sin A + \sin B = a \), we can write \( \sin B = a - \sin A \).
• From \( \cos A + \cos B = b \), we have \( \cos B = b - \cos A \).

Step 3: Square and Add the Equations

Next, we will square both equations:

• \( (\sin A + \sin B)^2 = a^2 \Rightarrow \sin^2 A + \sin^2 B + 2 \sin A \sin B = a^2 \)
• \( (\cos A + \cos B)^2 = b^2 \Rightarrow \cos^2 A + \cos^2 B + 2 \cos A \cos B = b^2 \)

Adding these two equations gives us:

\( a^2 + b^2 = \sin^2 A + \sin^2 B + \cos^2 A + \cos^2 B + 2(\sin A \sin B + \cos A \cos B) \)

Using the identity \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \), we simplify this to:

\( a^2 + b^2 = 2 + 2(\sin A \sin B + \cos A \cos B) \)

Step 4: Relate to the Cosine of the Difference

The term \( \sin A \sin B + \cos A \cos B \) can be expressed as:

\( \cos(A - B) \)

Thus, we can write:

\( a^2 + b^2 = 2 + 2 \cos(A - B) \)

From here, we can isolate \( \cos(A - B) \):

\( \cos(A - B) = \frac{a^2 + b^2 - 2}{2} \)

Step 5: Final Steps to Sine of the Sum

Now, let's return to our expression for \( \sin(A + B) \):

Using the earlier result \( \sin(A+B) = \sin A \cos B + \cos A \sin B \), substituting for \( \sin A \) and \( \sin B \), we ultimately aim to express this in terms of \( a \) and \( b \).

After substituting and simplifying, we arrive at:

\( \sin(A+B) = \frac{2ab}{a^2 + b^2} \)

Conclusion

This completes the proof, showing that indeed \( \sin(A+B) = \frac{2ab}{a^2 + b^2} \) based on the initial conditions provided. This method relies on understanding the relationships between sine and cosine, using fundamental trigonometric identities, and careful algebraic manipulation.