Question

Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin(\frac{1}{2} \tan^{-1} \frac{3}{4})$. If $S = \{\alpha \in R : y^2 = 1-x\}$, then $\sum_{\alpha \in S} 16\alpha^3$ is equal to

Answer 1

50240/729

Answer 2

Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin(\frac{1}{2} \tan^{-1} \frac{3}{4})$.

We express $x$ in terms of $\alpha$. Let $\theta = \tan^{-1} \alpha$. Then $\tan \theta = \alpha$. $x = \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\alpha}{1+\alpha^2}$.

Next, we calculate $y^2$. Let $\phi = \tan^{-1} \frac{3}{4}$. Using the principal value, $\phi \in (0, \frac{\pi}{2})$. Thus $\tan \phi = \frac{3}{4}$. We can construct a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is $\sqrt{3^2 + 4^2} = 5$. From this, $\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$. Now we use the half-angle identity for sine: $\sin^2(\frac{\phi}{2}) = \frac{1 - \cos \phi}{2}$. $y^2 = \sin^2(\frac{1}{2} \tan^{-1} \frac{3}{4}) = \sin^2(\frac{\phi}{2}) = \frac{1 - \frac{4}{5}}{2} = \frac{\frac{5-4}{5}}{2} = \frac{\frac{1}{5}}{2} = \frac{1}{10}$.

The condition given is $y^2 = 1-x$. Substituting the expressions for $x$ and $y^2$: $\frac{1}{10} = 1 - \frac{2\alpha}{1+\alpha^2}$. Rearranging the equation to solve for $\alpha$: $\frac{2\alpha}{1+\alpha^2} = 1 - \frac{1}{10} = \frac{9}{10}$. Multiply both sides by $10(1+\alpha^2)$: $20\alpha = 9(1+\alpha^2)$ $20\alpha = 9 + 9\alpha^2$ $9\alpha^2 - 20\alpha + 9 = 0$.

Let the roots of this quadratic equation be $\alpha_1$ and $\alpha_2$. The set $S$ consists of these roots, $S = \{\alpha_1, \alpha_2\}$. We need to find the value of $\sum_{\alpha \in S} 16\alpha^3 = 16\alpha_1^3 + 16\alpha_2^3 = 16(\alpha_1^3 + \alpha_2^3)$. From Vieta's formulas, for the quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $\alpha_1 + \alpha_2 = -\frac{b}{a}$ and the product of the roots is $\alpha_1 \alpha_2 = \frac{c}{a}$. For $9\alpha^2 - 20\alpha + 9 = 0$: Sum of roots: $\alpha_1 + \alpha_2 = -(\frac{-20}{9}) = \frac{20}{9}$. Product of roots: $\alpha_1 \alpha_2 = \frac{9}{9} = 1$.

We use the identity $\alpha_1^3 + \alpha_2^3 = (\alpha_1 + \alpha_2)^3 - 3\alpha_1 \alpha_2 (\alpha_1 + \alpha_2)$. Substituting the values of the sum and product of the roots: $\alpha_1^3 + \alpha_2^3 = \left(\frac{20}{9}\right)^3 - 3(1)\left(\frac{20}{9}\right)$ $\alpha_1^3 + \alpha_2^3 = \frac{8000}{729} - \frac{60}{9}$ To subtract the fractions, find a common denominator, which is 729. $\frac{60}{9} = \frac{60 \times 81}{9 \times 81} = \frac{4860}{729}$. $\alpha_1^3 + \alpha_2^3 = \frac{8000}{729} - \frac{4860}{729} = \frac{8000 - 4860}{729} = \frac{3140}{729}$.

Finally, we calculate the required sum: $\sum_{\alpha \in S} 16\alpha^3 = 16(\alpha_1^3 + \alpha_2^3) = 16 \times \frac{3140}{729}$. $16 \times 3140 = 50240$. So, $\sum_{\alpha \in S} 16\alpha^3 = \frac{50240}{729}$.

The final answer is $\boxed{\frac{50240}{729}}$.

Explanation of the solution:

1. Express $x = \sin(2 \tan^{-1} \alpha)$ as $\frac{2\alpha}{1+\alpha^2}$.
2. Calculate $y^2 = \sin^2(\frac{1}{2} \tan^{-1} \frac{3}{4}) = \frac{1 - \cos(\tan^{-1} \frac{3}{4})}{2} = \frac{1 - 4/5}{2} = \frac{1}{10}$.
3. Substitute $x$ and $y^2$ into the equation $y^2 = 1-x$: $\frac{1}{10} = 1 - \frac{2\alpha}{1+\alpha^2}$.
4. Solve the resulting equation for $\alpha$: $\frac{2\alpha}{1+\alpha^2} = \frac{9}{10} \implies 9\alpha^2 - 20\alpha + 9 = 0$.
5. Let the roots of this quadratic equation be $\alpha_1$ and $\alpha_2$. Use Vieta's formulas to find the sum and product of the roots: $\alpha_1 + \alpha_2 = \frac{20}{9}$ and $\alpha_1 \alpha_2 = 1$.
6. Calculate the sum of cubes $\alpha_1^3 + \alpha_2^3$ using the identity $(\alpha_1 + \alpha_2)^3 - 3\alpha_1 \alpha_2 (\alpha_1 + \alpha_2)$. $\alpha_1^3 + \alpha_2^3 = (\frac{20}{9})^3 - 3(1)(\frac{20}{9}) = \frac{8000}{729} - \frac{60}{9} = \frac{8000 - 4860}{729} = \frac{3140}{729}$.
7. Calculate the required sum $16(\alpha_1^3 + \alpha_2^3) = 16 \times \frac{3140}{729} = \frac{50240}{729}$.

The final answer is $\boxed{\frac{50240}{729}}$.

Note: A similar question has the value $\frac{3}{4}$ replaced by $\frac{4}{3}$. This leads to a much simpler integer answer of 130. It is possible that the given question had a typo, but the solution provided is for the question as written.