Solving for tan(alpha) in Inverse Trigonometric Sum
The problem requires finding the value of $\tan \alpha$, where $\alpha$ is defined by the equation:
$ \alpha = \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) $We will simplify the summation term first.
Simplifying the Inverse Tangent Sum
Recall the inverse tangent subtraction identity:
$ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right) $We can rewrite the term inside the summation:
$ \frac{2^{p-1}}{1 + 2^{2p-1}} = \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} $Here, let $x = 2^p$ and $y = 2^{p-1}$. Applying the identity, the summation term becomes:
$ \tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) $The summation is a telescoping series:
$ \sum_{p=1}^{11} (\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})) $Expanding the sum:
- $p=1: \tan^{-1}(2^1) - \tan^{-1}(2^0)$
- $p=2: \tan^{-1}(2^2) - \tan^{-1}(2^1)$
- ...
- $p=11: \tan^{-1}(2^{11}) - \tan^{-1}(2^{10})$
After cancellation, the sum simplifies to:
$ \tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \tan^{-1}(1) $Calculating alpha
Substitute the simplified sum back into the equation for $\alpha$:
$ \alpha = \frac{\pi}{4} + (\tan^{-1}(2048) - \tan^{-1}(1)) $Since $\frac{\pi}{4} = \tan^{-1}(1)$, we have:
$ \alpha = \tan^{-1}(1) + \tan^{-1}(2048) - \tan^{-1}(1) $ $ \alpha = \tan^{-1}(2048) $Finding tan(alpha)
The final step is to find the tangent of $\alpha$:
$ \tan \alpha = \tan(\tan^{-1}(2048)) $ $ \tan \alpha = 2048 $Similar Questions
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Let $\vec{a_k} = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k - 1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a_k}|^2}{\sum_{k=1}^n |\vec{b_k}|^2}$ is _____.
- If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
- Let the maximum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n} \pi^2$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to __________.
- The angle of elevation of the top $P$ of a vertical tower for a person standing at a point $A$ on the horizontal ground due north of the tower is $45^\circ$. Another person $B$ is standing $50$ m west of $A$ on the ground. If the angle of elevation of $P$ at $B$ is $30^\circ$ then the height (in meters) of the tower is
- Let $S = \{\theta \in [0,2\pi] - (\frac{\pi}{2}, \frac{3\pi}{2}) \mid \sin^{-1}(\sin\theta) + \cos^{-1}(\cos\theta) + \tan^{-1}(\tan\theta) = \frac{4\pi}{5}\}$ Then $\frac{30}{\pi} \sum_{\theta \in S} \theta$ is equal to ________
If $y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right)$, then $(x-y)^2+3y^2$ is equal to ____________.
Let A(4, -2), B(1, 1) and C(9, -3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ______________.
- If $\text{A} = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ and $\text{B} = \tan 81^\circ - \tan 3^\circ$, then $\frac{\text{B}}{\text{A}}$ is equal to _____.
Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
- The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
Let $\vec{a_k} = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k - 1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a_k}|^2}{\sum_{k=1}^n |\vec{b_k}|^2}$ is _____.
- If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
- Let the maximum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m}{n} \pi^2$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to __________.
