# Solving the Double Angle Tangent Formula for $\tan x$

The double angle formula for $$\tan(x)$$ is as follows: $$\tan(2x) = \frac{2\tan(x)}{1-\tan^2 (x)}$$ I wanted to see if I could solve this equation for $$\tan(x)$$—I figured that I could manipulate this equation to put it in the form of a quadratic equation**. $$\tan(2x)(\tan x)^2 + 2(\tan x) - \tan(2x) = 0 \\ \implies \tan(x) = \frac{-2 \pm \sqrt{4 - 4(\tan(2x))(-\tan(2x))}}{2\tan(2x)}$$ Conveniently, the expression for $$\tan(x)$$ simplifies to $$\tan(x) = \frac{-1 \pm \sec(2x)}{\tan(2x)}$$ Before calling it a day, I checked to see if any of these branches of the solution were extraneous. As it turns out, the negative branch is extraneous, and is actually equal to $$\tan{\left( x - \frac{\pi}{2} \right)} = -\cot(x)$$.

This is where I’m confused. Both branches of the expression are valid solutions to both the quadratic equation and the original double angle equation. So why isn’t $$\tan (x)$$ equal to both of them? I know that would be ridiculous, but I can’t see where this phase shift by $$\frac{\pi}{2}$$ comes from.

**I doubted at first whether the quadratic equation applies in a case where the coefficients (a, b, and c) are functions of the equation's independent variable, in this case $$x$$. However, I decided to continue anyway, since the alternative "safer" way to solve this would be to complete the square, but that's essentially equivalent to using the quadratic formula anyway.

• Any reason not to use the half-angle identity? It gives a somewhat different solution: $$\tan x = \csc 2x - \cot 2x = \frac{\tan 2x}{1+ \sec 2x}$$ Commented Jan 25, 2023 at 23:58
• I suppose it's notable that $$\frac{\tan 2x}{1+\sec 2x} = \frac{\sec 2x -1}{\tan 2x}$$ which is your final answer. I suspect the phase shift of $\pi/2$ comes from the fact that $\tan(2x - \pi) = \tan 2x$, since the tangent has a period of $\pi$ rather than $2\pi$. Edit: also notable that the standard identity for secant in terms of tangent is $\sec x = \pm \sqrt{1 + \tan x}$, meaning you don't need $\pm \sec x$ in your final expression. Commented Jan 26, 2023 at 0:18
• @EricSnyder That latter fact of yours explains why the solution is extraneous! That answers part of the question—that fact about the phase shift is useful too. Although, I'm still curious as to why the extraneous solution happens to be just a phase shift of $\tan x$. Commented Jan 26, 2023 at 3:25
• @EricSnyder By the way, the tanx within the square root should be squared. Commented Feb 11, 2023 at 17:23

If $$T$$ is the tangent of the double angle,

$$T=\frac{2t}{1-t^2},~ Tt^2+2t-T=0,~$$

$$~ t_1=\frac{\sqrt{1+T^2}-1}{T},~ t_2=\frac{-\sqrt{1+T^2}-1}{T},~$$

which can be also expressed in terms of secant you have given.

In the solution of a quadratic equation both solutions are correct.

Product of roots is $$-1$$ so that the vectors of single angles are orthogonal.

Sketched is for a situation when tangent double angle $$150^{\circ}$$ is

$$\tan 2\theta=\frac{-1}{\sqrt 3},\tan \theta_{1,2}=\pm(2-\sqrt 3),~$$

and for half angles

$$75^{\circ}, -15^{\circ}.~$$

Two solutions out of four in other quadrants viz.,

$$165^{\circ}, 255^{\circ}~$$

are not included in the sketch for clarity.

$$\tan\left(x - \frac{\pi}{2}\right) = \frac{-1}{\tan(x)} ~: ~0 < x < \pi/2.$$

Evaluating:

$$\frac{2\left(\frac{-1}{\tan(x)}\right)}{1 - \left[ ~\left(\frac{-1}{\tan(x)}\right)^2 ~\right]}$$

$$= \frac{\frac{-2}{\tan(x)}}{\frac{\tan^2(x) - 1}{\tan^2(x)}} = \frac{2\tan(x)}{1 - \tan^2(x)} = \tan(2x).$$

• This doesn't answer the question. I know that $\tan{\left (x - \frac{\pi}{2} \right)} = \frac{-1-\sec 2x}{\tan 2x}$ is a solution to the original equation, but I'm trying to figure out why the expression does not equate to $\tan (x)$ in the end. Commented Jan 26, 2023 at 3:17
• @user2661923 Saying that the $\tan 2x$ formula wasn't meant to be taken in reverse is... odd. The identity is an expression relating two functions, not the inverse of a function. That expression can, in fact, be reversed, and manipulated in many ways. Commented Jan 26, 2023 at 4:47
• @EricSnyder Suppose that $\tan(x) = A$ and $\tan(2x)=B.$ All I meant was that while this does imply that $~\displaystyle B = \frac{2A}{1 - A^2},~$ it does not (necessarily) imply that $~A~$ is the only real number $~x~$ such that $~\displaystyle B = \frac{2x}{1 - x^2}.~$ Commented Jan 26, 2023 at 4:59
• @EricSnyder Actually, I wasn't going that deeply. I (intentionally) avoided trying to conjure intuition on why the extraneous root happened to be a phase shift. My analysis is merely intended to show that the value of $~\displaystyle \frac{-1}{\tan(x)}~$ happens to be a root that is not (in general) equal to $~\tan(x)~$ and therefore must be an extraneous root. Then, I focused on the fact that it is typical for a quadratic equation to have two (unequal) roots. Commented Jan 26, 2023 at 5:08
• @Mailbox: At the risk of merely repeating the assertion ... The quadratic relates $\tan x$ and $\tan2x$. As a bonus, it relates $-\cot x$ and $\tan2x$ in exactly the same way. There's no reason to expect the two values to match. ... BTW: If you think of $\tan x$ as the slope of a line, then $-\cot x$ is the negative-reciprocal slope. So, the quadratic encodes some common quality of perpendicular lines. We see this particular ambiguity when seeking the axes of a conic from its second-degree eqn; that pesky $\pm$ is what distinguishes the major/transverse axis from the minor/conjugate.
– Blue
Commented Jan 26, 2023 at 5:30

Thanks to @EricSynder, @user2661923, and @Blue for your help in answering this question.

To begin, to answer the question of why this expression for $$\tan(x)$$ comes up, it turns out it's my fault. At the following step: $$\tan(x) = \frac{-1 \pm \sqrt{1 + \tan^2 (2x)}}{\tan(2x)}$$ The $$\pm \sqrt{1 + \tan^2 (2x)}$$ term should have simplified to $$+\sec(2x)$$, not $$\pm \sec(2x)$$. Otherwise, that would imply that $$\sec(x) = \sqrt{1 + \tan^2 (x)}$$ for all real numbers, but this equation is only correct for half of the $$\sec(x)$$ function (see a visual representation of this here.

Before I mention the phase shift, I should first regard @user2661923's point from their answer. Although $$\tan(x) \neq \frac{-1-\sec(2x)}{\tan(2x)} = -\cot(x)$$, $$-\cot(x)$$ is still a solution to the general quadratic. If, instead of the original tangent double-angle equation, we replace $$\tan(x)$$ with a general $$f(x)$$:

$$\tan(2x) = \frac{2f(x)}{1-(f(x))^2}$$

One could easily check that both $$f(x) = \tan(x)$$ and $$f(x) = -\cot(x)$$ are valid solutions to this equation (they are also both solutions to the implied quadratic). This is why @user2661923 did not like the use of the word "erroneous" when referring to the second solution, $$f(x) = -\cot(x)$$, since it is not so when thought about in this context.

To continue, let's consider the solution,$$f$$ ,to the quadratic that results from rearranging the above equation: $$\tan(2x)(f(x))^2 + 2(f(x)) - \tan(2x) = 0$$ $$\iff f(x) = \frac{-1 \pm \sqrt{1+\tan^2(2x)}}{\tan(2x)}$$

Note that $$f$$ has two branches, a positive one and a negative one. Let's analyze the individual branches of $$f(x)$$; to do so, it's best to see $$f(x)$$ plotted (link to the graph I made of $$f$$ on Desmos):

The positive branch of $$f$$ is shown in blue, and its negative branch is shown in green. Now, let's compare this to a similar graph, where I simply overlay the graphs of $$\tan(x)$$ and $$-\cot(x)$$ on top of each other:

Here, $$\tan(x)$$ is in purple, and $$-\cot(x)$$ is in black.

Notice: although the two graphs are composed of different functions, the two graphs as a whole are exactly the same! In other words, the individual branches that they are composed of are different, but the "sum" of the branches is the same nonetheless! This honestly blew my mind the first time I saw this.

This explains, in a way, why both $$\tan(x)$$ and $$-\cot(x)$$ come up when solving that equation. When overlayed on top of each other, the combined function that they make is identical to the technical solution to the original equation.

I wish I could provide a more complete solution with a more algebraic approach, but this was the best that I could do. If anyone else comes across this post and has any additional insight, I would love to read it.