4 Sine Pi over 10 by Cosine Pi over 5
Theorem
- $4 \sin \dfrac \pi {10} \cos \dfrac \pi 5 = 1$
Proof 1
\(\displaystyle \paren {z + 1} \paren {z^2 - 2 z \cos \dfrac \pi 5 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 5 + 1}\) | \(=\) | \(\displaystyle z^5 + 1\) | Complex Algebra Examples: $z^5 + 1$ | ||||||||||
\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {1 + i} \paren {i^2 - 2 i \cos \dfrac \pi 5 + 1} \paren {i^2 - 2 i \cos \dfrac {3 \pi} 5 + 1}\) | \(=\) | \(\displaystyle i^5 + 1\) | putting $z \gets i$ | |||||||||
\(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {1 + i} \paren {-1 - 2 i \cos \dfrac \pi 5 + 1} \paren {-1 - 2 i \cos \dfrac {3 \pi} 5 + 1}\) | \(=\) | \(\displaystyle i + 1\) | Definition of Imaginary Unit | |||||||||
\(\displaystyle -4 \paren {1 + i} \cos \dfrac \pi 5 \cos \dfrac {3 \pi} 5\) | \(=\) | \(\displaystyle i + 1\) | simplifying | ||||||||||
\(\displaystyle -4 \cos \dfrac \pi 5 \cos \dfrac {3 \pi} 5\) | \(=\) | \(\displaystyle 1\) | equating real parts | ||||||||||
\(\displaystyle -4 \cos \dfrac \pi 5 \cos \paren {\dfrac \pi {10} + \dfrac \pi 2}\) | \(=\) | \(\displaystyle 1\) | |||||||||||
\(\displaystyle -4 \cos \dfrac \pi 5 \paren {-\sin \dfrac \pi {10} }\) | \(=\) | \(\displaystyle 1\) | |||||||||||
\(\displaystyle 4 \cos \dfrac \pi 5 \sin \dfrac \pi {10}\) | \(=\) | \(\displaystyle 1\) |
$\blacksquare$
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.
\(\displaystyle 4 \sin \dfrac \pi {10} \cos \dfrac \pi 5\) | \(=\) | \(\displaystyle 1\) | 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 |
$\blacksquare$