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]
