$\tan2x$ in terms of $\cos x$ alone

Question

This is not exactly a homework question but something I was trying to do to get my basics back on track.

I wanted to find $\tan2x$ in terms of $\cos x$ alone. I was able to do it in terms of $\sin x$ alone.

$\tan2x = \sin2x/\cos2x$

Since, $\cos2x = 1-2\sin^2x$

Therefore, $\tan2x = (\sin2x / 1-2\sin^2x)$

Is it possible to do it in terms of $\cos x$ alone ?

Note that $\cos 2x = 1 - 2\sin^2x$ not $1 - \sin^2x$. – Michael Albanese Dec 29 '13 at 13:15 — Michael Albanese, Dec 29 '13 at 13:15
@MichaelAlbanese Yes, Sorry ! :) – Kartik Anand Dec 29 '13 at 13:16 — Kartik Anand, Dec 29 '13 at 13:16

Adi Dani · Answer 1 · 2013-12-29T13:51:01.667

5

$$\tan2x = \frac{\sin2x}{\cos2x}=\pm\frac{2\sin x\cos x}{\cos^2x-\sin^2x}=\pm\frac{2\cos x\sqrt{1-\cos^2x}}{2\cos^2x-1}$$

edited Dec 29 '13 at 13:51

answered Dec 29 '13 at 13:16

Adi Dani

16,949

Are the square roots necessary? :) – Kartik Anand Dec 29 '13 at 13:22
@KartikAnand. Yes. – Adi Dani Dec 29 '13 at 13:29
I meant is there are solution without square roots? – Kartik Anand Dec 29 '13 at 13:31
If $x=-\pi/6$, then $\tan2x=-\tan(\pi/3)=-\sqrt{3}$, while the expression on the right is $\sqrt{3}$. So your relation is false. – egreg Dec 29 '13 at 13:45
$\sin^2x+\cos^2x=1\Rightarrow\sin x=\pm\sqrt{1-\cos^2x}$ – Adi Dani Dec 29 '13 at 13:50
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The $\pm$ is of course meaningless. – egreg Dec 29 '13 at 13:51
Is from your opinion formula for solving quadratic equation $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ meaningless? – Adi Dani Dec 29 '13 at 13:55
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$\tan 2x$ has a unique value; you have to tell which one to choose, or the formula means nothing. – egreg Dec 29 '13 at 14:08
That I left for you. – Adi Dani Dec 29 '13 at 14:16
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@egreg: Knowing $\cos x$ itself determines $x$ only up to its sign, so the $\pm$ is essential. There will be two different possible values of $\tan 2x$. – hmakholm left over Monica Dec 29 '13 at 15:12
@HenningMakholm Of which only one is correct. – egreg Dec 29 '13 at 15:19
@egreg: How do you propose to distinguish between them? – hmakholm left over Monica Dec 29 '13 at 15:21
@HenningMakholm I did in my answer. – egreg Dec 29 '13 at 15:22
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@egreg: Only by assuming that you know $x$ which you don't. (If you knew $x$ you could just ignore the $\cos x$ and compute $\tan 2x$ directly). – hmakholm left over Monica Dec 29 '13 at 15:25

egreg · Accepted Answer · 2013-12-29T15:32:46.737

3

You can, but only using care. It's not restrictive to assume $x\in[0,2\pi]$, because $\tan2x$ has $2\pi$ as period. Then you have $$ \tan2x= \begin{cases} \dfrac{2\cos x\sqrt{1-\cos^2x}}{2\cos^2-1} & x\in[0,\pi]\\[2ex] -\dfrac{2\cos x\sqrt{1-\cos^2x}}{2\cos^2-1} & x\in[\pi,2\pi] \end{cases} $$ (the values $\pi/4$, $3\pi/4$, $5\pi/4$ and $7\pi/4$ are excluded, of course).

In other words, you can only express $|\tan2x|$ in terms of $\cos x$ alone.

edited Dec 29 '13 at 15:32

answered Dec 29 '13 at 14:02

egreg

238,574

your answer is more complete than my. +1 – Adi Dani Dec 29 '13 at 14:12
This assumes that you can ask whether $x$ is in one interval or another. But you can't do that when all we know about $x$ is its cosine,. – hmakholm left over Monica Dec 29 '13 at 15:22
@HenningMakholm Of course. So what? The information would be insufficient, which means you can't express $\tan2x$ in terms of $\cos x$ only. – egreg Dec 29 '13 at 15:23
So your answer is not an answer. The question is to determine $\tan 2x$ given only $\cos x$. Your answer requires information that the question says we don't have. – hmakholm left over Monica Dec 29 '13 at 15:24
@HenningMakholm Can you walk to the moon? I guess you can't. If the tools you have don't allow for getting an answer, the answer can't be obtained. – egreg Dec 29 '13 at 15:26

score 1 · Answer 3 · answered Dec 29 '13 at 15:13

Yes, it is possible to do it without radicals, BUT only if you relax the demands of the problem to include $\cos(x+\phi)$ as well:

$$\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\sin x\cos x}{\cos^2x-\sin^2x}=\frac{2\cos x\cos(x-\frac\pi2)}{2\cos^2x-1}$$

This way, you can also forget about worrying about signs! :-)

$\tan2x$ in terms of $\cos x$ alone

3 Answers3