Theorem: 4 Sine Pi over 10 by Cosine Pi over 5

:$4 \sin \dfrac \pi {10} \cos \dfrac \pi 5 = 1$

Proof:
Rewrite the {{LHS}}:
{{begin-eqn}}
{{eqn | l = 4 \sin \dfrac \pi {10} \cos \dfrac \pi 5
      | r = 1
      | c = {{hypothesis}}
}}
{{eqn | l = 4 \sin \dfrac \pi {10} \cos \dfrac {2 \pi} {10}
      | r = 1
      | c = multiplying the angle inside the cosine by $\dfrac 2 2$
}}
{{eqn | l = 4 \sin \dfrac \pi {10} \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}
      | r = 1
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | l = 4 \sin \paren { \dfrac 1 2 \dfrac \pi {5} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}
      | r = 1
      | c = factoring a $\dfrac 1 2$ out of $\dfrac \pi {10}$
}}
{{eqn | l = 4 \sqrt { \dfrac { 1 - \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1} } 2 } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}
      | r = 1
      | c = $\sin \dfrac \theta 2 = \sqrt { \dfrac { 1 - \cos \theta } 2 }$
}}
{{eqn | l = 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}
      | r = 1
      | c = simplification
}}
{{end-eqn}}
We then solve for $\theta = \dfrac \pi {10}$:
{{begin-eqn}}
{{eqn | l = \cos \dfrac \pi 2
      | r = 0
      | c = Cosine of Right Angle: $\dfrac \pi 2 = 5 \theta$
}}
{{eqn | l = \cos \paren { \dfrac \pi 5 + \dfrac {3 \pi} {10} }
      | r = 0
      | c = rewriting $5 \theta$ as $2 \theta + 3 \theta$
}}
{{eqn | l = \cos \dfrac \pi 5 \cos \dfrac {3\pi} {10} - \sin \dfrac \pi 5 \sin \dfrac {3\pi} {10}
      | r = 0
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | l = \paren { 2 \cos^2 \dfrac \pi {10} - 1 } \paren { 4 \cos^3 \dfrac {3\pi} {10} - 3 \cos \dfrac \pi {10} } - \paren { 2 \sin \dfrac \pi {10} \cos \dfrac \pi {10} } \paren { 3 \sin \dfrac \pi {10} - 4 \sin^3 \dfrac \pi {10} }
      | r = 0
      | c = Multiple-angle formulas (cosine, sine)
}}
{{eqn | l = \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \sin^2 \dfrac \pi {10} \cos \dfrac \pi {10} - 8 \sin^4 \dfrac \pi {10} \cos \dfrac \pi {10} }
      | r = 0
      | c = Multiplication
}}
{{eqn | l = \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \cos \dfrac \pi {10} - 6 \cos^3 \dfrac \pi {10} - 8 \cos \dfrac \pi {10} + 16 \cos^3 \dfrac \pi {10} - 8 \cos^5 \dfrac \pi {10} }
      | r = 0
      | c = Pythagorean Identity
}}
{{eqn | l = 16 \cos^5 \dfrac \pi {10} - 20 \cos^3 \dfrac \pi {10} + 5 \cos \dfrac \pi {10}
      | r = 0
      | c = Trivial simplification
}}
{{end-eqn}}
Having realized that the above process is the same as the derivation of the expansion formula for $\cos {5 \theta}$, we can generalize this expression and then plug in $\dfrac \pi {10}$.
{{begin-eqn}}
{{eqn | l = 16u^5 - 20u^3 + 5u
      | r = 0
      | c = $u = \cos \theta$
}}
{{eqn | l = u(16u^4 - 20u^2 + 5)
      | r = 0
      | c = Factoring a $u$ out of the polynomial
}}
{{eqn | l = 16u^4 - 20u^2 + 5
      | r = 0
      | c = Dividing both sides by $u$
}}
{{end-eqn}}
We were able to divide off the $u$ from both sides of that last line because we know that $\cos \dfrac \pi {10}$ cannot possibly equal $0$. From here, we use the quadratic formula to solve for $u^2$.
{{begin-eqn}}
{{eqn | l = u^2
      | r = \dfrac { - \paren { -20 } ± \sqrt { \paren { -20 }^2 - 4 \paren { 16 } \paren 5} } { 2 \paren { 16 } }
      | c =
}}
{{eqn | l = u^2
      | r = \dfrac { 20 ± \sqrt { 80 } } {32}
      | c =
}}
{{eqn | l = u^2
      | r = \dfrac { 5 ± \sqrt 5 } 8
      | c =
}}
{{end-eqn}}
Taking the square root of both sides, we can solve for $u$.
{{begin-eqn}}
{{eqn | l = u^2
      | r = \dfrac { 5 ± \sqrt 5 } 8
      | c =
}}
{{eqn | l = u
      | r = ± \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}
      | c =
}}
{{end-eqn}}
We know that $\dfrac \pi {10}$ is in the first quadrant, so we can change the exterior $±$ sign to a $+$ sign:
{{begin-eqn}}
{{eqn | l = u 
      | r = \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}
      | c =
}}
{{end-eqn}}
We know that $\dfrac \pi {10}$ is a lot closer to $0$ than to $\dfrac \pi 2$:
$\dfrac \pi {10} - 0 < \dfrac \pi 2 - \dfrac \pi {10}$
Therefore, the larger of the possible interior quantities of the radical should be correct.
{{begin-eqn}}
{{eqn | l = u
      | r = \dfrac 1 {\sqrt 8} \sqrt {5 + \sqrt 5}
      | c =
}}
{{eqn | l = u
      | r = \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }
      | c =
}}
{{end-eqn}}
Now that we have evaluated $\cos \dfrac \pi {10}$, we can substitute it for the most simplified expression from above with only $\cos \dfrac \pi {10}$:
{{begin-eqn}}
{{eqn | l = 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}
      | r = 1
      | c = Derived from original theorem
}}
{{eqn | l = 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { 2 \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 - 1}
      | r = 1
      | c = $u = \cos \dfrac \pi {10} = \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }$
}}
{{eqn | l = 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}
      | r = 1
      | c = $2 \paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 4 {16} a = \dfrac 1 4 a$
}}
{{eqn | l = 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}
      | r = 1
      | c = $\paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 2 {16} a = \dfrac 1 8 a$
}}
{{eqn | l = 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - 1}
      | r = 1
      | c = Multiplication
}}
{{eqn | l = 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - \dfrac 4 4}
      | r = 1
      | c = Common denominators
}}
{{eqn | l = 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 - 4 } 4}
      | r = 1
      | c = Combining fractions
}}
{{eqn | l = 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 1 + \sqrt 5 } 4}
      | r = 1
      | c = simplification
}}
{{eqn | l = \dfrac 4 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = Associative property
}}
{{eqn | l = \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = simplification
}}
{{eqn | l = \sqrt { \dfrac 8 8 - \dfrac { 5 + \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = Common denominators
}}
{{eqn | l = \sqrt { \dfrac { 8 - 5 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = combining fractions
}}
{{eqn | l = \sqrt { \dfrac { 3 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = simplification
}}
{{eqn | l = \sqrt { \dfrac { 6 - 2 \sqrt 5 } {16} } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = multiplying the radicand by $\dfrac 2 2$
}}
{{eqn | l = \sqrt { \dfrac { 1 - 2 \sqrt 5 + 5 } {16} } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = rewriting $k ± 2 \sqrt u$ as $\paren { k - u } ± 2 \sqrt u + u$ because $k - u + u = k$
}}
{{eqn | l = \sqrt { \dfrac { \paren { 1 - \sqrt 5 }^2 } {16} } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = rewriting the quantity as a square
}}
{{eqn | l = ± \dfrac 1 4 \paren { 1 - \sqrt 5 } \paren { 1 + \sqrt 5 }
      | r = 1
      | c = Square root of the radicand
}}
{{eqn | l = ± \dfrac 1 4 \paren { 1 - 5 }
      | r = 1
      | c = Multiplying $\paren { 1 - \sqrt 5 }$ and $\paren { 1 + \sqrt 5 }$
}}
{{eqn | l = ± \dfrac 1 4 × -4
      | r = 1
      | c = simplification
}}
{{eqn | l = ∓ 1
      | r = 1
      | c = Multiplication
}}
{{eqn | l = 1
      | r = 1
      | c = Sign rectification
}}
{{end-eqn}}
{{qed}}