Question

If \(A, B, C\) are the angles of a triangle, then \(\cos B + \cos C - \cos A + 1\) is equal to

• A. \(4\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\)
• B. \(-4\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\)
• C. \(4\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)
• D. \(4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\)

Hint:

Always use \(A+B+C=\pi\) in triangle trig problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a triangle, \( A + B + C = \pi \). We use sum-to-product identities to transform the expression.
Step 2: Detailed Explanation:
Consider the expression \( E = (\cos B + \cos C) + (1 - \cos A) \).
1. Use \( \cos B + \cos C = 2\cos(\frac{B+C}{2})\cos(\frac{B-C}{2}) \).
Since \( \frac{B+C}{2} = \frac{\pi - A}{2} = \frac{\pi}{2} - \frac{A}{2} \), then \( \cos(\frac{B+C}{2}) = \sin(\frac{A}{2}) \).
So, \( \cos B + \cos C = 2\sin(A/2)\cos(\frac{B-C}{2}) \).
2. Use \( 1 - \cos A = 2\sin^2(A/2) \).
Substituting both back into E: \[ E = 2\sin(A/2)\cos(\frac{B-C}{2}) + 2\sin^2(A/2) \] Factor out \( 2\sin(A/2) \): \[ E = 2\sin(A/2) [\cos(\frac{B-C}{2}) + \sin(A/2)] \] Replace \( \sin(A/2) \) with \( \cos(\frac{B+C}{2}) \): \[ E = 2\sin(A/2) [\cos(\frac{B-C}{2}) + \cos(\frac{B+C}{2})] \] Using \( \cos X + \cos Y = 2\cos(\frac{X+Y}{2})\cos(\frac{X-Y}{2}) \): \[ E = 2\sin(A/2) [ 2\cos(\frac{B}{2})\cos(\frac{C}{2}) ] \] \[ E = 4\sin(A/2)\cos(B/2)\cos(C/2) \].
Step 3: Final Answer:
The expression equals \( 4\sin(A/2)\cos(B/2)\cos(C/2) \).