Answer :

Sure! Let's find the algebraic expression for \(\cos \left(\frac{1}{2} \arccos x\right)\), given that \(x > 0\).

### Step-by-Step Solution

1. Understanding the Problem:
We need to express \(\cos \left(\frac{1}{2} \arccos x\right)\) in terms of \(x\), where \(x > 0\).

2. Using a Trigonometric Identity:
We can use the half-angle identity for cosine. One such identity is:
[tex]\[ \cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}} \][/tex]

3. Applying the Identity:
In this case, we can set \(\theta = \arccos x\).
Therefore, we have:
[tex]\[ \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{1 + \cos (\arccos x)}{2}} \][/tex]

4. Simplifying Inside the Square Root:
Recall that \(\cos (\arccos x) = x\). This is true because the arccosine function is the inverse of the cosine function.
Thus, we substitute \(x\) for \(\cos (\arccos x)\):
[tex]\[ \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{1 + x}{2}} \][/tex]

5. Final Algebraic Expression:
The expression simplifies to:
[tex]\[ \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{1 + x}{2}} \][/tex]

Or, written slightly differently:
[tex]\[ \cos \left(\frac{1}{2} \arccos x\right) = \sqrt{\frac{x}{2} + \frac{1}{2}} \][/tex]

Therefore, the algebraic expression for \(\cos \left(\frac{1}{2} \arccos x\right)\) when \(x > 0\) is:
[tex]\[ \sqrt{\frac{x}{2} + \frac{1}{2}} \][/tex]

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