Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ I have to simplify and evaluate this : 
$$\cot 70^\circ+4\cos 70^\circ$$
On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .
I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
 A: First $70=90-20$
We can express all in terms of $\cos(20)$ and use that $\frac{1}{2}=\cos(60)=4\cos^3(20)-3\cos(20)$.
Let's write $x=\cos(20), y=\sin(20)$ to write less.
So, $4x^3-3x-\frac{1}{2}=0$.
We square your expression such that we don't have to write radicals, but we can go without it too if we wanted.
$$\begin{align}
\left(\cot(70)+4\cos(70)\right)^2&=\left(\frac{\cos(90-20)+4\cos(90-20)\sin(90-20)}{\sin(90-20)}\right)^2\\
&=\left(\frac{y+4xy}{x}\right)^2\\
&=y^2\frac{16x^2+8x+1}{x^2}\\&=??\end{align}$$
But
$$\frac{1}{x}=8x^2-6.$$
$$\begin{align}??&=(1-x^2)(16x^2+8x+1)(8x^2-6)^2\\&=-1024 x^8-512 x^7+2496 x^6+1280 x^5-1952 x^4-1056 x^3+444 x^2+288 x+36\end{align}$$
Now divide this polynomial by $4x^3-3x-1/2$, which is zero. 
$$??=(-256 x^5-128 x^4+432 x^3+192 x^2-180 x-66)\cdot(4 x^3-3 x-1/2) + 3$$
The remainder gives you the value $3$. 
We could have worked too with $3\cdot(70)=270-60$ to get smaller numbers in the coefficients, but well, I already wrote it with $20$. You can try this technique with $\cos(70)$ directly to check that you understood how it works.
A: Multiply by $\sin70$ and you want to show $$\cos70+4\cos70\sin70=\sqrt{3}\sin70=2\sin60\sin70$$
A: The easy way :   
$$\dfrac{\cos70^\circ+4\cos70^\circ\sin70^\circ}{\sin70^\circ}  = \dfrac{\sin20^\circ+2\sin40^\circ}{\cos20^\circ} = \dfrac{\sin20^\circ+\sin40^\circ+\cos50^\circ}{\cos20^\circ}    $$
Sum to product formula gives :  
$$\dfrac{\cos10^\circ+\cos50^\circ}{\cos20^\circ} $$  
Again using the formula gives $2\cos30^\circ=\sqrt3\approx1.732$
A: Like Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $,
If $2\sin3x=-1$
$$\cot x+4\cos x=4\cos x-2\sin3x\cot x$$
$$=\dfrac{4\cos x\sin x-2\sin3x\cos x}{\sin x}$$
$$=\dfrac{2\sin2x-(\sin2x+\sin4x)}{\sin x}$$
$$=-2\cos3x\text{ as }\sin x\ne0$$
Here $x=70^\circ$
Generalization: 
$$\sin3x=-\dfrac12\implies3x=180^\circ n+(-1)^n(-30^\circ)\text{ where $n$ is any integer}$$
$\implies x=60^\circ n+(-1)^n(-10^\circ)$  where $n\equiv-1,0,1\pmod3$
$n=-1\implies x=-50^\circ$
$n=0\implies x=-10^\circ$
$n=1\implies x=70^\circ$
