I need to prove that: $$1+\tan x \tan 2x = \sec 2x.$$

I started this by making sec 1/cos and using the double angle identity for that and it didn't work at all in any way ever. Not sure why I can't do that, but something was wrong.

Anyways I looked at the solutions manual and they magic out $$1 + \tan x \tan 2x = 1+\tan x\left(\frac{2 \tan x}{1-\tan ^2x}\right) $$ which I recognize as the double angle forumla sort of, I just don't understand why I can use that and how they magiced it into this. It is just too difficult to think of an equation in an equation in an equation. I tried working with this and got nowhere near the right answer.

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    $\begingroup$ Since the right hand side does not contain "2x" (but rather, just "x"), you will have to use a double angle identity. I know it seems like magic, but there are some very suggestive hints that you can learn to look for. Sadly, we don't have much time to learn before the test! $\endgroup$ – The Chaz 2.0 Jun 24 '11 at 1:36
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    $\begingroup$ Adam, more people than you realize need to take trig in college! (Even quite a few that studied trig in high school...) Anyway: Nothing to feel ashamed of! $\endgroup$ – Namaste Jun 24 '11 at 1:48

I always like to rewrite things in terms of $\cos$ and $\sin$ when trying to verify identities. I tried to write out the steps in as much detail as possible. $$ \begin{align*} 1+\tan(x)\tan(2x) &= 1+\frac{\sin(x)}{\cos(x)}\frac{\sin(2x)}{\cos(2x)}\\ &= 1+\frac{\sin(x)2\sin(x)\cos(x)}{\cos(x)\cos(2x)} \quad\text{using the double angle formula for }\sin(2x)\\ &= 1+\frac{2\sin^2(x)}{\cos(2x)}\quad\text{cancelling }\cos(x)\\ &= \frac{\cos(2x)+2\sin^2(x)}{\cos(2x)}\quad\text{getting a common denominator}\\ &= \frac{\cos^2(x)-\sin^2(x)+2\sin^2(x)}{\cos(2x)}\quad\text{using the identity for }\cos(2x)\\ &= \frac{\cos^2(x)+\sin^2(x)}{\cos(2x)}\\ &= \frac{1}{\cos(2x)}\\ &= \sec(2x) \end{align*} $$

  • $\begingroup$ If you cancel $cos(x)$ from top and bottom, you should really also check the case where $cos(x)=0$. $\endgroup$ – Mark Bennet Jun 24 '11 at 15:23

Using the double angle formulas for $\sin$ and $\cos$, we get $$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}.$$ Multiplying the top and bottom by $\dfrac{1}{\cos^2(x)}$, we get $$\tan(2x)=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\qquad\dfrac{2\sin(x)}{\cos(x)}\qquad}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}=\frac{2\tan(x)}{1-\tan^2(x)}.$$ It is a common trick in mathematics to multiply an expression by $\dfrac{A}{A}$, where $A$ is some clever choice of expresssion. Here we used $A=\dfrac{1}{\cos^2(x)}$. This obviously won't change the value, because $\dfrac{A}{A}=1$ and multiplying by 1 doesn't do anything, but it might collapse some things in what you started with to a form you know how to deal with.

  • $\begingroup$ I am confused tan2x = 2tana divided by 1-tan^2a $\endgroup$ – Adam Jun 24 '11 at 1:43
  • $\begingroup$ @Adam: That is the formula that I have derived in my answer. $\endgroup$ – Zev Chonoles Jun 24 '11 at 1:46
  • $\begingroup$ I am having trouble seeing the relation. My understand of basic math is around grade school level. I see what you did, I ahve absolutely zero idea how to derive this like of 30 or so identities but I do have a sheet of paper with them written down and have maybe 4 memorized. $\endgroup$ – Adam Jun 24 '11 at 1:47
  • $\begingroup$ @Adam: Are you familiar with the double-angle formulas for $\sin$ and $\cos$ that I used in the first step? Where in the answer do you get stuck, I will explain more on that step? $\endgroup$ – Zev Chonoles Jun 24 '11 at 1:49
  • $\begingroup$ I am just really bad at math, I have trouble with simple math problems. I keep doing everything wrong, I have done it thousands of times but I still mess up simple multiplying, factoring, and even adding and subtracting. That is my main problem, after that I am really not a creative person and I have trouble changing these problems to fit some goal that I have really little idea how to get to, I mostly just trial and error but with my incredibly common math errors this is a very, very (atleast 30 minutes a problem) lengthy process. $\endgroup$ – Adam Jun 24 '11 at 1:53

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