find the $\cos 40(2\cos 80-1)=?$

Question

find the $\cos 40(2\cos 80-1)=?$

My try :

$\cos 40(2\cos 80-1)=2\cos 40 \cos 80 -\cos40=2\cos40 (2\cos ^240-1)-\cos 40\\=4\cos^340-3\cos 40=\cos 3(40)=\cos (120)=-1/2$

I do not want to use the formula $4\cos^3x -3\cos x=\cos 3x$ . Now How to solve ?

Answer 1

$$ \cos(40°)\cdot(2\cdot \cos(80°)-1)\\ = \cos(40°)\cdot(4\cos^2(40°)-3)\\ = 4\cos^3(40°)-3\cos(40°)\\ = \cos(120°)\\ = -\frac{1}{2} $$

Just look how simple the solution is. Why would you want to not use this approach.
Unless you want to prove the triple angle identity from scratch.

Or $$ \cos(40°)\cdot(2\cdot \cos(80°)-1)\\ = 2\cdot \cos(40°) \cdot \cos(80°) - \cos(40°)\\ = \frac{2\cdot \sin(40°)\cdot \cos(40°) \cdot \cos(80°)}{\sin(40 °)} - \cos(40°)\\ = \frac{\sin(20°)}{2\sin(40°)} - \cos(40°)\\ = \frac{1}{4\cos(20°)} - \cos(40°)\\ = \frac{1-4\cos(40°)\cos(20°)}{4\cos(20°)}\\ $$

And this mess is so unnecessary that I let you continue if you wish so.