4 Sine Pi over 10 by Cosine Pi over 5/Proof 2

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Theorem

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


Proof

First, we must rewrite the left hand side so that it reaches a form that is able to be substituted for what we determine is true through identities.

\(\displaystyle 4 \sin \dfrac \pi {10} \cos \dfrac \pi 5\) \(=\) \(\displaystyle 1\) Given
\(\displaystyle 4 \sin \dfrac \pi {10} \cos \dfrac {2 \pi} {10}\) \(=\) \(\displaystyle 1\) multiplying the angle inside the cosine by $\dfrac 2 2$
\(\displaystyle 4 \sin \dfrac \pi {10} \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\) \(=\) \(\displaystyle 1\) Double-angle formula (cosine)
\(\displaystyle 4 \sin \paren { \dfrac 1 2 \dfrac \pi {5} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\) \(=\) \(\displaystyle 1\) Factoring a $\dfrac 1 2$ out of $\dfrac \pi {10}$
\(\displaystyle 4 \sqrt { \dfrac { 1 - \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1} } 2 } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\) \(=\) \(\displaystyle 1\) $\sin \dfrac \theta 2 = \sqrt { \dfrac { 1 - \cos \theta } 2 }$
\(\displaystyle 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\) \(=\) \(\displaystyle 1\) Trivial simplification


Now that we have reduced the expression to just one trigonometric function of a single angle, we can use the fact that a multiple of the angle is $\dfrac \pi 2$, whose cosine is $0$, in order to solve for $\cos \dfrac \pi {10}$.

\(\displaystyle \cos \dfrac \pi 2\) \(=\) \(\displaystyle 0\) (New equation) Starting with $5 \theta$ and its value
\(\displaystyle \cos \paren { \dfrac \pi 5 + \dfrac {3\pi} {10} }\) \(=\) \(\displaystyle 0\) Rewriting $5 \theta$ as $2 \theta + 3 \theta$
\(\displaystyle \cos \dfrac \pi 5 \cos \dfrac {3\pi} {10} - \sin \dfrac \pi 5 \sin \dfrac {3\pi} {10}\) \(=\) \(\displaystyle 0\) Double-angle formula (cosine)
\(\displaystyle \paren { 2 \cos^2 \dfrac \pi {10} - 1 } \paren { 4 \cos^3 \dfrac {3\pi} {10} - 3 \cos \dfrac \pi {10} } - \paren { 2 \sin \dfrac \pi {10} \cos \dfrac \pi {10} } \paren { 3 \sin \dfrac \pi {10} - 4 \sin^3 \dfrac \pi {10} }\) \(=\) \(\displaystyle 0\) Multiple-angle formulas (cosine, sine)
\(\displaystyle \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \sin^2 \dfrac \pi {10} \cos \dfrac \pi {10} - 8 \sin^4 \dfrac \pi {10} \cos \dfrac \pi {10} }\) \(=\) \(\displaystyle 0\) Multiplication
\(\displaystyle \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \cos \dfrac \pi {10} - 6 \cos^3 \dfrac \pi {10} - 8 \cos \dfrac \pi {10} + 16 \cos^3 \dfrac \pi {10} - 8 \cos^5 \dfrac \pi {10} }\) \(=\) \(\displaystyle 0\) Pythagorean Identity
\(\displaystyle 16 \cos^5 \dfrac \pi {10} - 20 \cos^3 \dfrac \pi {10} + 5 \cos \dfrac \pi {10}\) \(=\) \(\displaystyle 0\) Trivial simplification


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 $\dfrac \pi {10}$.

\(\displaystyle 16u^5 - 20u^3 + 5u\) \(=\) \(\displaystyle 0\) $u = \cos \theta$
\(\displaystyle u(16u^4 - 20u^2 + 5)\) \(=\) \(\displaystyle 0\) Factoring a $u$ out of the polynomial
\(\displaystyle 16u^4 - 20u^2 + 5\) \(=\) \(\displaystyle 0\) Dividing both sides by $u$


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 $u^2$.

\(\displaystyle u^2\) \(=\) \(\displaystyle \dfrac { - \paren { -20 } ± \sqrt { \paren { -20 }^2 - 4 \paren { 16 } \paren 5} } { 2 \paren { 16 } }\)
\(\displaystyle u^2\) \(=\) \(\displaystyle \dfrac { 20 ± \sqrt { 80 } } {32}\)
\(\displaystyle u^2\) \(=\) \(\displaystyle \dfrac { 5 ± \sqrt 5 } 8\)


Taking the square root of both sides, we can solve for $u$.

\(\displaystyle u^2\) \(=\) \(\displaystyle \dfrac { 5 ± \sqrt 5 } 8\)
\(\displaystyle u\) \(=\) \(\displaystyle ± \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}\)


We know that $\dfrac \pi {10}$ is in the first quadrant, so we can change the exterior $±$ sign to a $+$ sign:

\(\displaystyle u\) \(=\) \(\displaystyle \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}\)


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.

\(\displaystyle u\) \(=\) \(\displaystyle \dfrac 1 {\sqrt 8} \sqrt {5 + \sqrt 5}\)
\(\displaystyle u\) \(=\) \(\displaystyle \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }\)


Now that we have evaluated $\cos \dfrac \pi {10}$, we can substitute it for the most simplified expression from above with only $\cos \dfrac \pi {10}$:

\(\displaystyle 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\) \(=\) \(\displaystyle 1\) Derived from original theorem
\(\displaystyle 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { 2 \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 - 1}\) \(=\) \(\displaystyle 1\) $u = \cos \dfrac \pi {10} = \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }$
\(\displaystyle 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}\) \(=\) \(\displaystyle 1\) $2 \paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 4 {16} a = \dfrac 1 4 a$
\(\displaystyle 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}\) \(=\) \(\displaystyle 1\) $\paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 2 {16} a = \dfrac 1 8 a$
\(\displaystyle 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - 1}\) \(=\) \(\displaystyle 1\) Multiplication
\(\displaystyle 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - \dfrac 4 4}\) \(=\) \(\displaystyle 1\) Common denominators
\(\displaystyle 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 - 4 } 4}\) \(=\) \(\displaystyle 1\) Combining fractions
\(\displaystyle 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 1 + \sqrt 5 } 4}\) \(=\) \(\displaystyle 1\) Trivial simplification
\(\displaystyle \dfrac 4 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Associative property
\(\displaystyle \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Trivial simplification
\(\displaystyle \sqrt { \dfrac 8 8 - \dfrac { 5 + \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Common denominators
\(\displaystyle \sqrt { \dfrac { 8 - 5 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Combining fractions
\(\displaystyle \sqrt { \dfrac { 3 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Trivial simplification
\(\displaystyle \sqrt { \dfrac { 6 - 2 \sqrt 5 } {16} } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Multiplying the radicand by $\dfrac 2 2$
\(\displaystyle \sqrt { \dfrac { 1 - 2 \sqrt 5 + 5 } {16} } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Rewriting $k ± 2 \sqrt u$ as $\paren { k - u } ± 2 \sqrt u + u$ because $k - u + u = k$
\(\displaystyle \sqrt { \dfrac { \paren { 1 - \sqrt 5 }^2 } {16} } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Rewriting the quantity as a square
\(\displaystyle ± \dfrac 1 4 \paren { 1 - \sqrt 5 } \paren { 1 + \sqrt 5 }\) \(=\) \(\displaystyle 1\) Square root of the radicand
\(\displaystyle ± \dfrac 1 4 \paren { 1 - 5 }\) \(=\) \(\displaystyle 1\) Multiplying $\paren { 1 - \sqrt 5 }$ and $\paren { 1 + \sqrt 5 }$
\(\displaystyle ± \dfrac 1 4 × -4\) \(=\) \(\displaystyle 1\) Trivial simplification
\(\displaystyle ∓ 1\) \(=\) \(\displaystyle 1\) Multiplication
\(\displaystyle 1\) \(=\) \(\displaystyle 1\) Sign rectification

$\blacksquare$