If `x=sin(2tan^(-1)3)and y=sin((1)/(2)tan^(-1)(4/3))`, then

Options

A. `2x=1-y`
B. `x^(2)=1-2y`
C. `x^(2)=1+y`
D. `y^(2)=2x-1`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) given by the equations:

\[ x = \sin(2 \tan^{-1}(3)) \]
\[ y = \sin\left(\frac{1}{2} \tan^{-1}\left(\frac{4}{3}\right)\right) \]

### Step 1: Calculate \( x \)

Let \( \theta = \tan^{-1}(3) \). Then, we have:

\[ x = \sin(2\theta) \]

Using the double angle formula for sine, we know:

\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]

Now, we need to find \( \sin(\theta) \) and \( \cos(\theta) \).

Since \( \theta = \tan^{-1}(3) \), we can represent it in a right triangle where the opposite side is 3 and the adjacent side is 1. The hypotenuse can be calculated using the Pythagorean theorem:

\[ \text{Hypotenuse} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]

Thus, we have:

\[ \sin(\theta) = \frac{3}{\sqrt{10}}, \quad \cos(\theta) = \frac{1}{\sqrt{10}} \]

Now substituting these values back into the double angle formula:

\[ x = 2 \cdot \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

### Step 2: Calculate \( y \)

Let \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \). Then, we have:

\[ y = \sin\left(\frac{1}{2} \phi\right) \]

Using the half angle formula for sine, we know:

\[ \sin\left(\frac{1}{2} \phi\right) = \sqrt{\frac{1 - \cos(\phi)}{2}} \]

Now, we need to find \( \cos(\phi) \).

Since \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \), we can represent it in a right triangle where the opposite side is 4 and the adjacent side is 3. The hypotenuse can be calculated using the Pythagorean theorem:

\[ \text{Hypotenuse} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Thus, we have:

\[ \cos(\phi) = \frac{3}{5} \]

Now substituting this value back into the half angle formula:

\[ y = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Step 3: Check the options

Now we have:

\[ x = \frac{3}{5}, \quad y = \frac{1}{\sqrt{5}} \]

We will check the given options:

1. **Option 1:** \( 2x = 1 - y \)

\[ 2 \cdot \frac{3}{5} = \frac{6}{5}, \quad 1 - \frac{1}{\sqrt{5}} \text{ (not equal)} \]

2. **Option 2:** \( x^2 = 1 - 2y \)

\[ \left(\frac{3}{5}\right)^2 = \frac{9}{25}, \quad 1 - 2 \cdot \frac{1}{\sqrt{5}} \text{ (not equal)} \]

3. **Option 3:** \( x^2 = 1 + y \)

\[ \frac{9}{25} = 1 + \frac{1}{\sqrt{5}} \text{ (not equal)} \]

4. **Option 4:** \( x^2 = 2x - 1 \)

\[ \frac{9}{25} = 2 \cdot \frac{3}{5} - 1 \implies \frac{9}{25} = \frac{6}{5} - 1 = \frac{6}{5} - \frac{5}{5} = \frac{1}{5} = \frac{5}{25} \text{ (not equal)} \]

### Conclusion

After checking all options, we find that none of the options are satisfied.