What does $\cos(2\tan^{−1} x)$ mean and how do solve?

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Solution 1

In words, you are starting with a number $x$, which you can think of as a length if $x$ is positive.

The next thing is to find the arctangent of that $x$, which is denoted $\tan^{-1}(x)$. You could think of this as putting a 1-meter stick pointing out from your foot and then raising a pole that is $x$ meters tall from the other end. The angle from your foot to the top of the pole is the arctangent of $x$.

Once you have this angle, you are asked to double it, and so you still have an angle, but one that is twice as big. Proceeding with the construction, you build a ramp from your foot to the top of the pole. You lay down a $1$-meter stick up this ramp from your foot. And then you attach another $x$-meter pole from the end of this second meter stick, extending out from the ramp perpendicular to it. The doubled angle is now the angle at your foot between the ground and the tip of the second pole.

And lastly you are asked to find the cosine of the doubled angle. Tie string from your foot to the tip of the second pole, and then attach a third meter stick from the tip of your foot alongside the string. Locate the end of this third meter stick, and shoot straight down to the ground. How far is this point from your foot? That distance is $\cos\left(2\tan^{-1}(x)\right)$.


To compute it, it's a general fact that $\cos(2A)=\left(\cos(A)\right)^2-\left(\sin(A)\right)^2$. So this means $$\begin{align} \cos(2\tan^{-1}(x))&=\left(\cos\left(\tan^{-1}(x)\right)\right)^2-\left(\sin\left(\tan^{-1}(x)\right)\right)^2 \end{align}$$

Considering a right triangle with legs of length $1$ and $x$, then the angle opposite the leg of length $x$ is $\tan^{-1}(x)$, almost by definition. So then the cosine of that angle is $\frac{1}{\sqrt{1+x^2}}$ and the sine of that angle is $\frac{x}{\sqrt{1+x^2}}$. So

$$\begin{align} \cos(2\tan^{-1}(x))&=\left(\frac{1}{\sqrt{1+x^2}}\right)^2-\left(\frac{x}{\sqrt{1+x^2}}\right)^2\\ &=\frac{1-x^2}{1+x^2} \end{align}$$

Solution 2

The $x$ in your expression denotes a certain real number, unknown to us. At any rate $-\infty<x<\infty$. So $x=\tan\alpha$ for a certain angle $\alpha$ with $-{\pi\over2}<\alpha<{\pi\over2}$. This means that $\tan^{-1}x$, which is another way of writing $\arctan x$, is nothing but this $\alpha$, and $2\tan^{-1}x$ is $2\alpha$, an angle between $-\pi$ and $\pi$.

The question ask us to express the quantity $Q:=\cos(2\alpha)$ in terms of $x$, i.e., in terms of $\tan\alpha$. Now $$\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha={\cos^2\alpha-\sin^2\alpha\over \cos^2\alpha+\sin^2\alpha}={1-\tan^2\alpha\over 1+\tan^2\alpha}\ ,$$ so that we finally obtain $$Q={1-x^2\over1+x^2}\ .$$

Solution 3

For $\cos (2 \tan ^{-1}x)$; the precedence of operations is:

  1. Use $x$ to look up the $\arctan$ table (if you still have one) or find the result from your calculator by pressing (shift, tan, x, EXE) depending on what machine you are using.

  2. $2$ means double your finding in step 1.

  3. Apply $\cos$ to the result found in step 2

$\tan$ of an angle is used to find the ratio between the opposite side to the adjacent side.

Inverse of tangent is then the opposite operation. That is, if you know that ratio, you use the inverse of tangent (sometime it is written as $\arctan$, or $\tan^{-1}$ or shift tan for some calculator) to find out what that angle is.

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Updated on August 31, 2022

Comments

  • user137452
    user137452 about 1 year
    1. Why is the $2$ inside the parentheses with $\tan^{-1}x$? Is'nt it $\cos2$?

    2. What is the question really asking me? Find the inverse of $\tan$?

    3. What is the inverse of $2\tan^1x$?

    • goblin GONE
      goblin GONE over 9 years
      The concept of "solving" applies to equations $\cos (2 \tan^{-1} x) = 0$, not expressions like $\cos (2 \tan^{-1} x)$. I suggest editing the body of the question with an actual statement of what you're trying to do (to simplify, perhaps?) and changing the title appropriately.
  • user137452
    user137452 over 9 years
    I have to simplify cos(2tan−^1 x). What steps should I take w/o calculator.
  • Mick
    Mick over 9 years
    You have to know what x is first. Then step 1. Furthermore, the expression in your title is NOT an equation. Hence, it cannot be solved.