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Edit: The page which is marked to be "similar", the answers on there have absolutely no meaning to me. One of the answers uses this method which I don't like, another one assumes the solution and works it off on that, and another user uses a formula derived from another page. None of them showed work to any of these proofs, like I have stated before, I shouldn't have to hunt for an answer this basic, I am more than confident enough with the capabilities of people on this website, I do not think this question is that complex compared to other problems.

This question: $$\tan^{-1}(\tan(20^{\circ})\tan(40^{\circ})\tan(80^{\circ}))$$ was given in our Precalculus class. I have no idea where to even start, I've seen one solution which is quite arbitrary and insane and involves creating a formula for $2\cos(a)\cos(b)$ and $2\sin(a)\sin(b)$ when you write the tangents in terms of sine and cosine. The method involves the sum and difference formulas, and it is ridiculous because it feels like when you are doing this you already knew how to solve the problem to come up with such weird manners. I'm prone to believe that there are a multitude of other more realistic methods.

I've tried two methods so far, both are quite similar: substituting 20 degrees with u, using the double angle formula, and deriving the quadruple angle formula of course, the expression doesn't simplify nicely, and then taking an arctangent of this random expression isn't possible. The next thing I did was substituting 40 degrees as v and using the half and double angle formulas, again, it doesn't simplify nicely.

If anyone can use the methods I did above and expand on it, that would be great. Preferably, an answer without the use of calculus is preferred, but both are fine because I would like to see any solution really. (Please no hyperbolic trig, I do not understand that topic)

SovietWizard
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  • I looked at the answers and I didn't really understand, some of them looked at the answer and tried to prove it, but without knowing the answer, how can you solve it, that's what I was wondering. – SovietWizard Dec 17 '24 at 06:31
  • Excuse me, can you reopen the question again, I shouldn't have to hunt for the solutions?? If you had actually looked at the page, none of them answered in the way which I stated just above here, 2 of the answers used arbitrary formulas, without proof, and they seem like I said kind of absurd, 1 of those answers uses the formula I mentioned here which I said I don't want. The other 3rd answer assumed the solution and then worked it off on that. Close the question when it's properly answered If you didn't understand, English isn't my first language, so sorry. – SovietWizard Dec 17 '24 at 06:38
  • Often, posters will close questions because people will come on the site demanding us to do the work for them (or under the guise of 'I have an exam and need to have this question answered now') and because they showed no effort, the question got downvoted and then closed.

    Yours is different - you've shown the effort, but if you can tell us where you went wrong, we can help you. In fact, I will help reopen the question so you can get that help.

     – bjcolby15 Dec 17 '24 at 11:55
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    is the "insane" formula you refer to:

    $$2 \sin(a) \sin(b) = \cos(a-b)- \cos(a+b) $$

    because this formula is fairly common

     – Edmaand Vlad Phiyik Dec 17 '24 at 12:38
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    "How dare you not answer my question the way I want? You're doing it for nothing, so I expect better!" You might as well insist that we answer your question in iambic pentameter. – John Hughes Dec 17 '24 at 15:39

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