Given that A+B =120¢ª and CosA + Cos B = 1/¡î2 show that cos(A+B)/2 =1/¡î2
To prove that cos(A+B)/2 = 1/√2, we need to apply some trigonometric identities and algebraic manipulation. Let's start step by step:
1. Recall the double-angle formula for cosine:
cos(2θ) = 2cos²(θ) - 1
2. Express cos(A + B) using the double-angle formula:
cos(A + B) = cos(2(A + B)/2)
= 2cos²((A + B)/2) - 1
3. Substitute the given values of A + B and CosA + Cos B:
cos(A + B) = 2cos²(120°/2) - 1
= 2cos²(60°) - 1
= 2(1/2)² - 1
= 2(1/4) - 1
= 1/2 - 1
= 1/2 - 2/2
= -1/2
4. Now, we need to find cos(A + B)/2:
cos(A + B)/2 = cos((-1/2)/2)
= cos(-1/4)
5. Notice that the value of cos(-θ) is the same as cos(θ). Therefore,
cos(A + B)/2 = cos(1/4)
6. Find the value of cos(1/4) using a calculator or trigonometric tables. The decimal approximation is approximately 0.9659.
Hence, we can see that cos(A + B)/2 = 0.9659, which is not equal to 1/√2. Therefore, the given condition CosA + Cos B = 1/√2 does not satisfy the equation cos(A + B)/2 = 1/√2.