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I have taken the value of $\cos 20^o$ to be $x$.
Now, $$\cos 20^o\cos 40^o\cos 60^o\cos 80^o = \dfrac{\cos 20^o}{2}\cos(40^o)\cos(80^o)$$ $$=\dfrac{x}{2}\Bigg( \dfrac{\cos(40^o + 80^o) + \cos(40^o - 80^o)}{2} \Bigg)$$ $$=\dfrac{x}{2}\Bigg( \dfrac{\dfrac{-1}{\text{ }2} + \cos 40^o}{2} \Bigg) = \dfrac{x}{4}\Bigg( \dfrac{-1}{\text{ }2}+2x^2-1 \Bigg)$$ $$=\dfrac{x}{8}(4x^2-3) = \dfrac{4x^3-3x}{8}$$

Can this even be continued (I mean to ask if this method will lead me any further)? If yes, how do I continue it?

Thanks!

PS : Please don't provide alternatives, I want to try them myself first. This question is only for the sake of asking if this method can be continued and it if can be, then how.

Rajdeep Sindhu
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1 Answers1

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You are one step away from the final answer. Use the fact that $$\cos3x=4\cos^3x-3\cos x$$ and the problem is solved.

Alan
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  • Thank You! One soft-question though, should I memorize a few identities of $\sin(nx)$, $\cos(nx)$ for $n \in \Bbb N$ or just derive them at the time of use? I think I should go with the latter. – Rajdeep Sindhu Jul 03 '20 at 18:07
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    Trig identities for $n\leqslant 3$ often come in handy. I think you should remember them. You can always derive them if you forget though :) – Alan Jul 03 '20 at 19:13