Find the value of :
` sin^() (2 tan^(-1). 1/4) +cos (tan^(-1)2sqrt(2))`

Answer

To solve the expression \( \sin(2 \tan^{-1}(\frac{1}{4})) + \cos(\tan^{-1}(2\sqrt{2})) \), we will break it down into manageable steps.

### Step 1: Calculate \( \sin(2 \tan^{-1}(\frac{1}{4})) \)

Using the double angle formula for sine:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Let \( \theta = \tan^{-1}(\frac{1}{4}) \). Then:
\[ \tan(\theta) = \frac{1}{4} \implies \text{opposite} = 1, \text{adjacent} = 4 \]
Using the Pythagorean theorem, we find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{1^2 + 4^2} = \sqrt{17} \]
Now, we can find \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \sin(\theta) = \frac{1}{\sqrt{17}}, \quad \cos(\theta) = \frac{4}{\sqrt{17}} \]
Now substituting back into the double angle formula:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \cdot \frac{1}{\sqrt{17}} \cdot \frac{4}{\sqrt{17}} = \frac{8}{17} \]

### Step 2: Calculate \( \cos(\tan^{-1}(2\sqrt{2})) \)

Let \( \phi = \tan^{-1}(2\sqrt{2}) \). Then:
\[ \tan(\phi) = 2\sqrt{2} \implies \text{opposite} = 2\sqrt{2}, \text{adjacent} = 1 \]
Using the Pythagorean theorem, we find the hypotenuse:
\[ \text{hypotenuse} = \sqrt{(2\sqrt{2})^2 + 1^2} = \sqrt{8 + 1} = 3 \]
Now, we can find \( \cos(\phi) \):
\[ \cos(\phi) = \frac{1}{3} \]

### Step 3: Combine the results

Now we can combine the results from Step 1 and Step 2:
\[ \sin(2 \tan^{-1}(\frac{1}{4})) + \cos(\tan^{-1}(2\sqrt{2})) = \frac{8}{17} + \frac{1}{3} \]
To add these fractions, we need a common denominator. The least common multiple of 17 and 3 is 51:
\[ \frac{8}{17} = \frac{8 \times 3}{17 \times 3} = \frac{24}{51} \]
\[ \frac{1}{3} = \frac{1 \times 17}{3 \times 17} = \frac{17}{51} \]
Now, add the two fractions:
\[ \frac{24}{51} + \frac{17}{51} = \frac{41}{51} \]

### Final Answer
Thus, the value of \( \sin(2 \tan^{-1}(\frac{1}{4})) + \cos(\tan^{-1}(2\sqrt{2})) \) is:
\[ \boxed{\frac{41}{51}} \]