Show that: `cos(2tan^-1(1/7))=sin(4tan^-1(1/3))`

Explanation

To show that \( \cos(2\tan^{-1}(1/7)) = \sin(4\tan^{-1}(1/3)) \), we will evaluate both sides step by step.

### Step 1: Evaluate the Left-Hand Side (LHS)

We start with the left-hand side:

\[ \text{LHS} = \cos(2\tan^{-1}(1/7)) \]

Let \( \theta = \tan^{-1}(1/7) \). Then, we have:

\[ \tan(\theta) = \frac{1}{7} \]

Using the double angle formula for cosine:

\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]

### Step 2: Substitute the value of \( \tan(\theta) \)

Now, substituting \( \tan(\theta) = \frac{1}{7} \):

\[ \tan^2(\theta) = \left(\frac{1}{7}\right)^2 = \frac{1}{49} \]

Now substitute this into the cosine formula:

\[ \cos(2\theta) = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} = \frac{\frac{49 - 1}{49}}{\frac{49 + 1}{49}} = \frac{48/49}{50/49} \]

### Step 3: Simplify the expression

Now, simplifying gives:

\[ \cos(2\theta) = \frac{48}{50} = \frac{24}{25} \]

Thus, we have:

\[ \text{LHS} = \frac{24}{25} \]

### Step 4: Evaluate the Right-Hand Side (RHS)

Now we evaluate the right-hand side:

\[ \text{RHS} = \sin(4\tan^{-1}(1/3)) \]

Let \( \phi = \tan^{-1}(1/3) \). Then:

\[ \tan(\phi) = \frac{1}{3} \]

Using the double angle formula for sine:

\[ \sin(4\phi) = \sin(2 \cdot 2\phi) = 2\sin(2\phi)\cos(2\phi) \]

### Step 5: Calculate \( \sin(2\phi) \) and \( \cos(2\phi) \)

Using the formulas:

\[ \sin(2\phi) = \frac{2\tan(\phi)}{1 + \tan^2(\phi)} \quad \text{and} \quad \cos(2\phi) = \frac{1 - \tan^2(\phi)}{1 + \tan^2(\phi)} \]

Calculating \( \tan^2(\phi) \):

\[ \tan^2(\phi) = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

Now substituting into the sine and cosine formulas:

\[ \sin(2\phi) = \frac{2 \cdot \frac{1}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \frac{2}{3} \cdot \frac{9}{10} = \frac{6}{10} = \frac{3}{5} \]

\[ \cos(2\phi) = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} = \frac{\frac{8}{9}}{\frac{10}{9}} = \frac{8}{10} = \frac{4}{5} \]

### Step 6: Substitute back into \( \sin(4\phi) \)

Now substituting back into the equation for \( \sin(4\phi) \):

\[ \sin(4\phi) = 2 \cdot \sin(2\phi) \cdot \cos(2\phi) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \]

Thus, we have:

\[ \text{RHS} = \frac{24}{25} \]

### Conclusion

Since both sides are equal:

\[ \text{LHS} = \text{RHS} = \frac{24}{25} \]

Therefore, we conclude that:

\[ \cos(2\tan^{-1}(1/7)) = \sin(4\tan^{-1}(1/3)) \]