a) Given that A+B =120¢ª and cos A + cos B = 1/¡î2 show that cos(A+B)/2 =1/¡î2

To solve this trigonometric equation, we need to use some trigonometric identities and algebraic manipulations.

Let's start by using the double-angle identity for cosine: cos(2θ) = 2cos²θ - 1. Rearranging this equation, we get: cos²θ = (cos(2θ) + 1)/2.

Now, let's apply this identity to the first equation: A + B = 120°. We know that cos(A + B) = cos(120°). Using the double-angle identity, cos(120°) = 2cos²(60°) - 1. Simplifying further, we have cos(120°) = 2(1/2) - 1 = 0.

Next, let's use the sum-to-product identity for cosine: cos(A + B) = cosAcosB - sinAsinB. We can rewrite the given equation cos A + cos B = 1/√2 as cos A + cos B = cos(45°). Applying the sum-to-product identity, we have: cos(45°) = cosAcosB - sinAsinB.

Now, let's equate the two values of cos A + cos B that we have obtained:
0 = cosAcosB - sinAsinB.

From here, we need to solve for cos(A + B)/2, which is equal to √((1 + cos(2(A + B)))/2) using the half-angle identity for cosine.

Substituting the value of cos(2(A + B)) that we found earlier, we get:
cos(A + B)/2 = √((1 + 0)/2) = √(1/2) = 1/√2.

Thus, we have shown that cos(A + B)/2 = 1/√2.

Note: The double-angle identity, sum-to-product identity, and half-angle identity for cosine are fundamental trigonometric identities that should be memorized to help solve trigonometric equations like this one.