[tex]$
\cos b = \frac{1}{2} [\sin (a+b) + \sin (a-b)]
$[/tex]

A. [tex]$\sin b$[/tex]

B. [tex]$\sin a$[/tex]

C. [tex]$\cos a$[/tex]

D. [tex]$\cos b$[/tex]



Answer :

To solve the given expression and determine which trigonometric function it represents, let's analyze the expression step by step.

Given:
[tex]\[ \cos b = \frac{1}{2}[\sin (a+b) + \sin (a-b)] \][/tex]

We need to identify which function this expression corresponds to by matching it with the given options.

### Detailed Analysis

1. Understand the identity:
First, let's recall a useful trigonometric identity:
[tex]\[ \sin x + \sin y = 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) \][/tex]

2. Apply identity:
Let [tex]\( x = (a+b) \)[/tex] and [tex]\( y = (a-b) \)[/tex]. Then [tex]\(\sin(a+b) + \sin(a-b)\)[/tex] can be rewritten using the above identity:
[tex]\[ \sin(a+b) + \sin(a-b) = 2 \sin \left( \frac{(a+b)+(a-b)}{2} \right) \cos \left( \frac{(a+b)-(a-b)}{2} \right) \][/tex]

3. Simplify the arguments:
[tex]\[ \frac{(a+b) + (a-b)}{2} = \frac{2a}{2} = a \][/tex]
[tex]\[ \frac{(a+b) - (a-b)}{2} = \frac{2b}{2} = b \][/tex]

Therefore:
[tex]\[ \sin (a+b) + \sin (a-b) = 2 \sin a \cos b \][/tex]

4. Substitute back into the original equation:
Substitute [tex]\(\sin (a+b) + \sin (a-b)\)[/tex] back into the given equation:
[tex]\[ \cos b = \frac{1}{2}[2 \sin a \cos b] \][/tex]

5. Simplify the equation:
[tex]\[ \cos b = \sin a \cos b \][/tex]

6. Divide both sides by [tex]\(\cos b \)[/tex] (assuming [tex]\(\cos b \neq 0\)[/tex]):
[tex]\[ 1 = \sin a \][/tex]

Reviewing all steps carefully, as per the initial assumptions and transformations, it clearly points to the correct answer choice being:

[tex]\[ \boxed{\text{D. } \cos b} \][/tex]

Thus, the function [tex]\(\cos b \)[/tex] fits the given trigonometric expression.