Sum of Cosines of Four Times Angles of Triangle
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Theorem
Let $\triangle ABC$ be a triangle.
Then:
- $\cos 4 A + \cos 4 B + \cos 4 C = 4 \cos 2 A \cos 2 B \cos 2 C - 1$
Proof
First we note that:
| \(\ds A + B + C\) | \(=\) | \(\ds 180 \degrees\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 A + 2 B + 2 C\) | \(=\) | \(\ds 360 \degrees\) | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 2 A + 2 B\) | \(=\) | \(\ds 360 \degrees - 2 C\) |
That is, $2 C$ is the conjugate of $2 A + 2 B$.
Then:
| \(\ds \cos 4 A + \cos 4 B + \cos 4 C\) | \(=\) | \(\ds 2 \map \cos {2 A + 2 B} \map \cos {2 A - 2 B} + \cos 4 C\) | Cosine plus Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos {360 \degrees - 2 C} \map \cos {2 A - 2 B} + \cos 4 C\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 C \map \cos {2 A - 2 B} + \cos 4 C\) | Cosine of Conjugate Angle | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 C \paren {\cos 2 A \cos 2 B + \sin 2 A \sin 2 B} + \cos 4 C\) | Cosine of Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 C \paren {\cos 2 A \cos 2 B + \sin 2 A \sin 2 B} + \paren {2 \cos^2 2 C - 1}\) | Double Angle Formula for Cosine: Corollary $1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 A \cos 2 B \cos 2 C + 2 \paren {\sin 2 A \sin 2 B + \cos 2 C} \cos 2 C - 1\) | multiplying out and rearranging | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 A \cos 2 B \cos 2 C + 2 \paren {\sin 2 A \sin 2 B + \map \cos {360 \degrees - \paren {2 A + 2 B} } } \cos 2 C - 1\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 A \cos 2 B \cos 2 C + 2 \paren {\sin 2 A \sin 2 B + \map \cos {2 A + 2 B} } \cos 2 C - 1\) | Cosine of Conjugate Angle | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 A \cos 2 B \cos 2 C + 2 \paren {\sin 2 A \sin 2 B + \paren {\cos 2 A \cos 2 B - \sin 2 A \sin 2 B} } \cos C - 1\) | Cosine of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos 2 A \cos 2 B \cos 2 C + 2 \paren {\sin 2 A \sin 2 B - \sin 2 A \sin 2 B} \cos 2 C + 2 \cos 2 A \cos 2 B \cos 2 C - 1\) | multiplying out | |||||||||||
| \(\ds \) | \(=\) | \(\ds 4 \cos 2 A \cos 2 B \cos 2 C - 1\) | simplifying |
$\blacksquare$
Also presented as
This result can also be presented as:
- $\cos 4 A + \cos 4 B + \cos 4 C + 1 = 4 \cos 2 A \cos 2 B \cos 2 C$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercises $\text {XXXII}$: $20$.