Find the value of expression: `sin(2tan^(-1)(1/3))+cos(tan^(-1)2sqrt(2))`

Explanation

To find the value of the expression \( \sin(2\tan^{-1}(1/3)) + \cos(\tan^{-1}(2\sqrt{2})) \), we will follow a step-by-step approach.

### Step 1: Simplify \( \sin(2\tan^{-1}(1/3)) \)

We use the double angle formula for sine:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
Let \( \theta = \tan^{-1}(1/3) \). Then:
\[ \sin(2\tan^{-1}(1/3)) = 2\sin(\tan^{-1}(1/3))\cos(\tan^{-1}(1/3)) \]

### Step 2: Find \( \sin(\tan^{-1}(1/3)) \) and \( \cos(\tan^{-1}(1/3)) \)

Using the definition of tangent:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3} \]
We can form a right triangle where the opposite side is 1 and the adjacent side is 3. Using the Pythagorean theorem:
\[ \text{hypotenuse} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]
Thus, we have:
\[ \sin(\tan^{-1}(1/3)) = \frac{1}{\sqrt{10}}, \quad \cos(\tan^{-1}(1/3)) = \frac{3}{\sqrt{10}} \]

### Step 3: Substitute back into the sine expression

Now substituting these values back:
\[ \sin(2\tan^{-1}(1/3)) = 2 \cdot \frac{1}{\sqrt{10}} \cdot \frac{3}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

### Step 4: Simplify \( \cos(\tan^{-1}(2\sqrt{2})) \)

Let \( \alpha = \tan^{-1}(2\sqrt{2}) \). Then:
\[ \tan(\alpha) = 2\sqrt{2} \]
This implies a right triangle where the opposite side is \( 2\sqrt{2} \) and the adjacent side is 1. Using the Pythagorean theorem:
\[ \text{hypotenuse} = \sqrt{(2\sqrt{2})^2 + 1^2} = \sqrt{8 + 1} = \sqrt{9} = 3 \]
Thus:
\[ \cos(\tan^{-1}(2\sqrt{2})) = \frac{1}{3} \]

### Step 5: Combine the results

Now we can combine the results:
\[ \sin(2\tan^{-1}(1/3)) + \cos(\tan^{-1}(2\sqrt{2})) = \frac{3}{5} + \frac{1}{3} \]

### Step 6: Find a common denominator and simplify

The common denominator for \( 5 \) and \( 3 \) is \( 15 \):
\[ \frac{3}{5} = \frac{9}{15}, \quad \frac{1}{3} = \frac{5}{15} \]
Thus:
\[ \frac{3}{5} + \frac{1}{3} = \frac{9}{15} + \frac{5}{15} = \frac{14}{15} \]

### Final Answer

The value of the expression is:
\[ \boxed{\frac{14}{15}} \]