Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$

Solution 1:

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 $2\cos30^\circ=\sqrt3\approx1.732$

Solution 2:

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.