Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$

How to find the value of $$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$

manually ?

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What do you mean "manually", and what do you mean "find the value"? Express it in some way in terms of what? – Arturo Magidin Nov 17 '10 at 6:02
@Arturo Magidin:In mathematica,N[Sqrt[3] Cot[20 Degree] - 4 Cos[20 Degree]] = 1. I want to do this manually. – Quixotic Nov 17 '10 at 6:04

$$\displaystyle \sin(60^{\circ}-20^{\circ}) = \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$

$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$

$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - 2 \sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 20^{\circ}$$

Multiply by $\displaystyle \frac{2}{\sin 20^{\circ}}$

$$\displaystyle \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ} = 1$$

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Hey, thanks, but could you please tell me the way of thinking for this kind of problem ? – Quixotic Nov 17 '10 at 6:31
@Deb: $\sqrt{3}$ reminded me of thinking about $60^{\circ}$. Writing in terms of sin and cos gives sin 20 cos 20, which gets $40^{\circ}$ into the picture. – Aryabhata Nov 17 '10 at 6:34
Hm, thanks :-) Is there any way of developing this way of thinking ? or it's about practice ? – Quixotic Nov 17 '10 at 6:38
@Deb: Practice helps, especially if you don't give up too soon. – Aryabhata Nov 17 '10 at 6:43
@Debanjan: You should practice on the diagonalized square and the half right angled triangle. These triangles together with addition formulas helps a lot. – AD. Nov 17 '10 at 9:26

If $\cos3\theta=\frac12,$

$$\tan3\theta\cot\theta-4\cos\theta$$ $$=\frac{\sin3\theta\cos\theta}{\cos3\theta\sin\theta}-4\cos\theta$$ $$=\frac{\sin3\theta\cos\theta-4\cos3\theta\sin\theta\cos\theta}{\cos3\theta\sin\theta}$$

$$=\frac{\sin3\theta\cos\theta-\sin2\theta}{\frac12\sin\theta}\text { as }\sin2x=2\sin x\cos x\text{ and } \cos3\theta=\frac12$$

$$=\frac{\sin4\theta+\sin2\theta-2\sin2\theta}{\sin\theta}\text { applying } 2\sin A\cos B=\sin(A+B)+\sin(A-B)$$

$$=\frac{\sin4\theta-\sin2\theta}{\sin\theta}$$

$$=\frac{2\sin\theta\cos3\theta}{\sin\theta}\text { applying } \sin 2C-\sin 2D=2\sin(C-D)\cos(C+D)$$

$$=2\cos3\theta=1$$

Now, $$\cos3\theta=\frac12=\cos60^\circ\implies 3\theta=2n180^\circ\pm60^\circ\text{ where } n \text{ is any integer}$$

So, $\theta=(6n\pm1)20^\circ$ where $n$ is any integer

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Let $a\cos\theta+b\cot\theta=c$

$\implies a\cos\theta=c-b\cot\theta=\frac{c\sin\theta-b\cos\theta}{\sin\theta}$

$\implies a\cos\theta\sin\theta=c\sin\theta-b\cos\theta$

Putting $c=r\cos\alpha,b=r\sin\alpha$ where $r>0$

Squaring & adding we get $r^2=c^2+b^2\implies r=+\sqrt{b^2+c^2}$ and $\frac{\sin\alpha}b=\frac{\cos\alpha}c=\frac1r=\frac1{\sqrt{b^2+c^2}}$

$\implies c\sin\theta-b\cos\theta=\sqrt{b^2+c^2}\sin(\theta-\alpha)$ and $\cos\theta\sin\theta=\frac{\sin2\theta}2$

$$a\sin2\theta=2\sqrt{b^2+c^2}\sin(\theta-\alpha)$$

Now, the solution of $P\sin x= Q\sin A$ is in general intractable unless $P=0$ or $Q=0$ or $P=\pm Q\ne0$

Here

$1:$ if $a=0,$ the problem can be solved easily.

$2:$ if $b^2+c^2=0\implies b=c=0$ ( as $b,c$ are real), the problem can be solved easily.

$3:$ So, for non-trivial cases, either $a=2\sqrt{b^2+c^2}$ or $a=-2\sqrt{b^2+c^2}$

$$\begin{array}{|c|c|c|} \hline \text{ Case } & a=2\sqrt{b^2+c^2} & a=-2\sqrt{b^2+c^2} \\ \hline \text{ Comparison} & \sin2\theta=\sin(\theta-\alpha) & \sin2\theta=-\sin(\theta-\alpha)=\sin(\alpha-\theta),\text{ as }\sin(-x)=-\sin x \\ \hline \text{General Solution} & 2\theta=n180^\circ+(-1)^n(\theta-\alpha)\text{ where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(\alpha-\theta)\text{ where }n\text{ is any integer } \\ \hline n=2m & \alpha=m360^\circ-\theta\equiv-\theta\pmod{360^\circ} & \alpha=3\theta-m360^\circ\equiv3\theta \\ \hline n=2m+1& \alpha=3\theta-(2m+1)180^\circ\equiv 3\theta-180^\circ & \alpha=(2m+1)180^\circ-\theta\equiv180^\circ-\theta \\ \hline \end{array}$$

Here $a=-4,b=\sqrt3,\theta=20^\circ,$

So, $a=-2\sqrt{b^2+c^2}$ as $\sqrt{b^2+c^2} > 0$

$\implies\sqrt{b^2+c^2}=-\frac a2=2, \alpha=3\theta=60^\circ$ or $\alpha=180^\circ-\theta=160^\circ$

So, $\sin\alpha=\frac b{\sqrt{b^2+c^2}}=\frac{\sqrt3}2\implies \alpha=60^\circ$

So, $c=\cos\alpha\cdot \sqrt{b^2+c^2}=\frac12\cdot 2=1$

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Could somebody please help me to show the left most column which is invisible in the answer. – lab bhattacharjee Apr 23 '13 at 16:25

$$\frac{2\sin60^\circ\cdot\cos20^\circ-2(2\sin20^\circ\cos20^\circ)}{\sin20^\circ}$$

Using Werner Formula we get, $$\frac{\sin80^\circ+\sin40^\circ-2\sin40^\circ}{\sin20^\circ}$$

Using Prosthaphaeresis Formula, $\sin80^\circ-\sin40^\circ=2\sin20^\circ\cos60^\circ$

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Thanks for the edit and for linking it here. – antman Mar 28 '15 at 9:46