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

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.

• This looks like a job for trig identities! – Hurkyl Jun 28 '14 at 21:00

Hint: $$\cos 40^{\circ}=\cos (60^{\circ}-20^{\circ})=\sin 20^{\circ} \sin 60^{\circ}+\cos 20^{\circ} \cos 60^{\circ}=\frac{\sqrt{3}}{2}\sin 20^{\circ}+\frac{1}{2}\cos 20^{\circ}$$

• 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. – David H Jun 28 '14 at 21:19


\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}

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 $\displaystyle x=20^\circ$