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

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