4 Sine Pi over 10 by Cosine Pi over 5

Theorem

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

Proof 1

\(\ds \paren {z + 1} \paren {z^2 - 2 z \cos \dfrac \pi 5 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 5 + 1}\)

\(=\)

\(\ds z^5 + 1\)

Complex Algebra Examples: $z^5 + 1$

\(\ds \leadsto \ \ \)

\(\ds \paren {1 + i} \paren {i^2 - 2 i \cos \dfrac \pi 5 + 1} \paren {i^2 - 2 i \cos \dfrac {3 \pi} 5 + 1}\)

\(=\)

\(\ds i^5 + 1\)

putting $z \gets i$

\(\ds \leadsto \ \ \)

\(\ds \paren {1 + i} \paren {-1 - 2 i \cos \dfrac \pi 5 + 1} \paren {-1 - 2 i \cos \dfrac {3 \pi} 5 + 1}\)

\(=\)

\(\ds i + 1\)

Definition of Imaginary Unit

\(\ds -4 \paren {1 + i} \cos \dfrac \pi 5 \cos \dfrac {3 \pi} 5\)

\(=\)

\(\ds i + 1\)

simplifying

\(\ds -4 \cos \dfrac \pi 5 \cos \dfrac {3 \pi} 5\)

\(=\)

\(\ds 1\)

equating real parts

\(\ds -4 \cos \dfrac \pi 5 \cos \paren {\dfrac \pi {10} + \dfrac \pi 2}\)

\(=\)

\(\ds 1\)

\(\ds -4 \cos \dfrac \pi 5 \paren {-\sin \dfrac \pi {10} }\)

\(=\)

\(\ds 1\)

\(\ds 4 \cos \dfrac \pi 5 \sin \dfrac \pi {10}\)

\(=\)

\(\ds 1\)

$\blacksquare$

Proof 2

A proof that uses only simple trigonometry and algebra can be achieved, though it is longer.

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.

\(\ds 4 \sin \dfrac \pi {10} \cos \dfrac \pi 5\)

\(=\)

\(\ds 1\)

Given

\(\ds 4 \sin \dfrac \pi {10} \cos \dfrac {2 \pi} {10}\)

\(=\)

\(\ds 1\)

multiplying the angle inside the cosine by $\dfrac 2 2$

\(\ds 4 \sin \dfrac \pi {10} \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\)

\(=\)

\(\ds 1\)

Double-angle formula (cosine)

\(\ds 4 \sin \paren { \dfrac 1 2 \dfrac \pi {5} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\)

\(=\)

\(\ds 1\)

Factoring a $\dfrac 1 2$ out of $\dfrac \pi {10}$

\(\ds 4 \sqrt { \dfrac { 1 - \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1} } 2 } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\)

\(=\)

\(\ds 1\)

$\sin \dfrac \theta 2 = \sqrt { \dfrac { 1 - \cos \theta } 2 }$

\(\ds 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\)

\(=\)

\(\ds 1\)

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 $\cos \dfrac \pi {10}$.

\(\ds \cos \dfrac \pi 2\)

\(=\)

\(\ds 0\)

(New equation) Starting with $5 \theta$ and its value

\(\ds \cos \paren { \dfrac \pi 5 + \dfrac {3\pi} {10} }\)

\(=\)

\(\ds 0\)

Rewriting $5 \theta$ as $2 \theta + 3 \theta$

\(\ds \cos \dfrac \pi 5 \cos \dfrac {3\pi} {10} - \sin \dfrac \pi 5 \sin \dfrac {3\pi} {10}\)

\(=\)

\(\ds 0\)

Double-angle formula (cosine)

\(\ds \paren { 2 \cos^2 \dfrac \pi {10} - 1 } \paren { 4 \cos^3 \dfrac {3\pi} {10} - 3 \cos \dfrac \pi {10} } - \paren { 2 \sin \dfrac \pi {10} \cos \dfrac \pi {10} } \paren { 3 \sin \dfrac \pi {10} - 4 \sin^3 \dfrac \pi {10} }\)

\(=\)

\(\ds 0\)

Multiple-angle formulas (cosine, sine)

\(\ds \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \sin^2 \dfrac \pi {10} \cos \dfrac \pi {10} - 8 \sin^4 \dfrac \pi {10} \cos \dfrac \pi {10} }\)

\(=\)

\(\ds 0\)

Multiplication

\(\ds \paren { 8 \cos^5 \dfrac \pi {10} - 10 \cos^3 \dfrac \pi {10} + 3 \cos \dfrac \pi {10} } - \paren { 6 \cos \dfrac \pi {10} - 6 \cos^3 \dfrac \pi {10} - 8 \cos \dfrac \pi {10} + 16 \cos^3 \dfrac \pi {10} - 8 \cos^5 \dfrac \pi {10} }\)

\(=\)

\(\ds 0\)

Pythagorean Identity

\(\ds 16 \cos^5 \dfrac \pi {10} - 20 \cos^3 \dfrac \pi {10} + 5 \cos \dfrac \pi {10}\)

\(=\)

\(\ds 0\)

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 $\dfrac \pi {10}$.

\(\ds 16u^5 - 20u^3 + 5u\)

\(=\)

\(\ds 0\)

$u = \cos \theta$

\(\ds u(16u^4 - 20u^2 + 5)\)

\(=\)

\(\ds 0\)

Factoring a $u$ out of the polynomial

\(\ds 16u^4 - 20u^2 + 5\)

\(=\)

\(\ds 0\)

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 $u^2$.

\(\ds u^2\)

\(=\)

\(\ds \dfrac { - \paren { -20 } ± \sqrt { \paren { -20 }^2 - 4 \paren { 16 } \paren 5} } { 2 \paren { 16 } }\)

\(\ds u^2\)

\(=\)

\(\ds \dfrac { 20 ± \sqrt { 80 } } {32}\)

\(\ds u^2\)

\(=\)

\(\ds \dfrac { 5 ± \sqrt 5 } 8\)

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

\(\ds u^2\)

\(=\)

\(\ds \dfrac { 5 ± \sqrt 5 } 8\)

\(\ds u\)

\(=\)

\(\ds ± \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}\)

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

\(\ds u\)

\(=\)

\(\ds \dfrac 1 {\sqrt 8} \sqrt {5 ± \sqrt 5}\)

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.

\(\ds u\)

\(=\)

\(\ds \dfrac 1 {\sqrt 8} \sqrt {5 + \sqrt 5}\)

\(\ds u\)

\(=\)

\(\ds \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }\)

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

\(\ds 4 \sqrt { 1 - \cos^2 \dfrac \pi {10} } \paren { 2 \paren { \cos \dfrac \pi {10} }^2 - 1}\)

\(=\)

\(\ds 1\)

Derived from original theorem

\(\ds 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { 2 \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 - 1}\)

\(=\)

\(\ds 1\)

$u = \cos \dfrac \pi {10} = \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } }$

\(\ds 4 \sqrt { 1 - \paren { \dfrac 1 4 \sqrt { 2 \paren { 5 + \sqrt 5 } } }^2 } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}\)

\(=\)

\(\ds 1\)

$2 \paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 4 {16} a = \dfrac 1 4 a$

\(\ds 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac 1 4 \paren { 5 + \sqrt 5 } - 1}\)

\(=\)

\(\ds 1\)

$\paren { \dfrac 1 4 \sqrt { 2 \paren { a } } }^2 = \dfrac 2 {16} a = \dfrac 1 8 a$

\(\ds 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - 1}\)

\(=\)

\(\ds 1\)

Multiplication

\(\ds 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 } 4 - \dfrac 4 4}\)

\(=\)

\(\ds 1\)

Common denominators

\(\ds 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 5 + \sqrt 5 - 4 } 4}\)

\(=\)

\(\ds 1\)

Combining fractions

\(\ds 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { \dfrac { 1 + \sqrt 5 } 4}\)

\(=\)

\(\ds 1\)

Trivial simplification

\(\ds \dfrac 4 4 \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Associative property

\(\ds \sqrt { 1 - \dfrac 1 8 \paren { 5 + \sqrt 5 } } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Trivial simplification

\(\ds \sqrt { \dfrac 8 8 - \dfrac { 5 + \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Common denominators

\(\ds \sqrt { \dfrac { 8 - 5 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Combining fractions

\(\ds \sqrt { \dfrac { 3 - \sqrt 5 } 8 } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Trivial simplification

\(\ds \sqrt { \dfrac { 6 - 2 \sqrt 5 } {16} } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Multiplying the radicand by $\dfrac 2 2$

\(\ds \sqrt { \dfrac { 1 - 2 \sqrt 5 + 5 } {16} } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Rewriting $k ± 2 \sqrt u$ as $\paren { k - u } ± 2 \sqrt u + u$ because $k - u + u = k$

\(\ds \sqrt { \dfrac { \paren { 1 - \sqrt 5 }^2 } {16} } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Rewriting the quantity as a square

\(\ds ± \dfrac 1 4 \paren { 1 - \sqrt 5 } \paren { 1 + \sqrt 5 }\)

\(=\)

\(\ds 1\)

Square root of the radicand

\(\ds ± \dfrac 1 4 \paren { 1 - 5 }\)

\(=\)

\(\ds 1\)

Multiplying $\paren { 1 - \sqrt 5 }$ and $\paren { 1 + \sqrt 5 }$

\(\ds ± \dfrac 1 4 × -4\)

\(=\)

\(\ds 1\)

Trivial simplification

\(\ds ∓ 1\)

\(=\)

\(\ds 1\)

Multiplication

\(\ds 1\)

\(=\)

\(\ds 1\)

Sign rectification

$\blacksquare$