`cos(2tan^(-1)(1/2))=?`

Options

A. `3/5`
B. `4/5`
C. `7/8`
D. none of these

Explanation

To solve the problem \( \cos(2\tan^{-1}(1/2)) \), we can use the properties of inverse trigonometric functions. Here’s the step-by-step solution:

### Step 1: Use the Double Angle Identity
We know that:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
For \( \theta = \tan^{-1}(1/2) \), we can express \( \cos(2\tan^{-1}(1/2)) \) using the identity above.

### Step 2: Find \( \sin(\tan^{-1}(1/2)) \) and \( \cos(\tan^{-1}(1/2)) \)
From the definition of \( \tan^{-1} \):
\[ \tan(\theta) = \frac{1}{2} \]
This means that in a right triangle, the opposite side is 1 and the adjacent side is 2. We can find the hypotenuse using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{5} \]
Thus,
\[ \sin(\theta) = \frac{1}{\sqrt{5}} \quad \text{and} \quad \cos(\theta) = \frac{2}{\sqrt{5}} \]

### Step 3: Substitute into the Double Angle Formula
Now, substituting into the double angle formula:
\[ \cos(2\tan^{-1}(1/2)) = 2\cos^2(\theta) - 1 \]
Substituting \( \cos(\theta) \):
\[ \cos(2\tan^{-1}(1/2)) = 2\left(\frac{2}{\sqrt{5}}\right)^2 - 1 \]
Calculating \( \left(\frac{2}{\sqrt{5}}\right)^2 \):
\[ \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5} \]
Thus,
\[ \cos(2\tan^{-1}(1/2)) = 2 \cdot \frac{4}{5} - 1 = \frac{8}{5} - 1 = \frac{8}{5} - \frac{5}{5} = \frac{3}{5} \]

### Final Answer
Therefore, the value of \( \cos(2\tan^{-1}(1/2)) \) is:
\[ \boxed{\frac{3}{5}} \]