Cause I hate cos

We know that $\cos { 60° } =\frac { 1 }{ 2 }$ and the value of the rest of the expression becomes $\frac { 1 }{ 2 } \left( \frac { 1 }{ { 2 }^{ 3 }\sin { 20° } } \right) .\sin { \left( { 2 }^{ 3 }.\sin { 20 } ° \right) } \\ \Rightarrow \frac { 1 }{ 16 } \times \frac { 1 }{ \sin { 20° } } .\sin { 160° } \\ \Rightarrow \frac { 1 }{ 16\sin { 20° } } .\sin { (180°-20° } )\\ \Rightarrow \frac { 1 }{ 16 } \left( \frac { \sin { 20° } }{ \sin { 20° } } \right) =\frac { 1 }{ 16 }$ Hence the inverse of the given expression is ${ \left( \frac { 1 }{ 16 } \right) }^{ -1 }=16$

cosA.cos2A.cos4A.cos8A............... upto n terms

=sin(2 Greatest angle)/2^(Total number of terms) sin(Smallest angle)

In this case,

(sin2x80)/(2^4)(sin20)

=(sin160)/16sin20

=sin20/16sin20

=1/16

Answer = (1/16)^-1 = 16

Been a while since I used Chebyshev polynomials, so here we go.

We're working with CP of the first kind. Ignore the 60 for now so we have $\cos(20)\cos(40)\cos(80)$ . Our polynomial will be of degree three and have roots of $\cos(3\theta)$ . The first coefficient is 2 to the power of the degree of the equation-1.

$2^2x^3+...+0=\cos(3\cdot20)$

$4x^3+...-1/2=0$

The product of this polynomial's roots is the negation of the last coefficient divided by the first.

$-\dfrac{-1/2}{4}=\dfrac{1}{8}$ . Multiplying this by $\cos(60)$ we have our answer $\dfrac{1}{16}$