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I am trying to solve this task i.e. calculate this expression without using calculator, in terms of known values for angles such as 30,60,90,180 degrees :).

$$\frac{2\cos40^\circ-\cos20^\circ}{\sin20^\circ}$$

Thanks.

Edit: Special thanks to David H. The problem was, indeed, unsolvable, until I discovered a mistake in my textbook and corrected it. Thanks everyone.

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  • $\begingroup$ This looks like a job for trig identities! $\endgroup$ – Hurkyl Jun 28 '14 at 21:00
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Hint: $\cos 40^{\circ}=\cos (60^{\circ}-20^{\circ})=\sin 20^{\circ} \sin 60^{\circ}+\cos 20^{\circ} \cos 60^{\circ}=\frac{\sqrt{3}}{2}\sin 20^{\circ}+\frac{1}{2}\cos 20^{\circ} $

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    $\begingroup$ This hint is possibly misleading because it does not lead to a solution of the proposed problem. In fact, the proposed problem is technically unsolvable. $\endgroup$ – David H Jun 28 '14 at 21:19
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{2\cos\pars{40^{\circ}} - \cos\pars{20}^{\circ} \over \sin\pars{20}^{\circ}}: \ {\large ?}}$

\begin{align} &\color{#66f}{\large{2\cos\pars{40^{\circ}} - \cos\pars{20}^{\circ} \over \sin\pars{20}^{\circ}}} ={2\cos\pars{40^{\circ}}\cos\pars{60^{\circ}} - \cos\pars{20}^{\circ}\cos\pars{60^{\circ}} \over \sin\pars{20}^{\circ}\cos\pars{60^{\circ}}} \\[3mm]&={\cos\pars{40^{\circ}} -\ \overbrace{\bracks{\cos\pars{60^{\circ}}\cos\pars{20}^{\circ} + \sin\pars{60^{\circ}}\sin\pars{20^{\circ}}}}^{\ds{=\ \cos\pars{40^{\circ}}}}\ +\ \sin\pars{60^{\circ}}\sin\pars{20^{\circ}} \over \sin\pars{20}^{\circ}\cos\pars{60^{\circ}}} \\[3mm]&=\tan\pars{60^{\circ}} = \color{#66f}{\large\root{3}} \approx 1.7321 \end{align}

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As $\displaystyle\cos60^\circ=\frac12, I=2\cos(60^\circ-x)-\cos x=2\cos(60^\circ-x)-2\cos60^\circ\cos x$

Using Werner's formula, $\displaystyle I=2\cos(60^\circ-x)-[\cos(60^\circ-x)+\cos(60^\circ+x)]$

$\displaystyle\implies I=\cos(60^\circ-x)-\cos(60^\circ+x)=2\sin60^\circ\sin x$

Here $\displaystyle x=20^\circ$

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