Cosine of Difference/Proof 2
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Theorem
- $\map \cos {a - b} = \cos a \cos b + \sin a \sin b$
Proof
Consider two radii $OP$ and $OQ$ of a unit circle whose center is at the origin of a Cartesian plane.
Let:
\(\ds \angle xOP\) | \(=\) | \(\ds B\) | ||||||||||||
\(\ds \angle xOQ\) | \(=\) | \(\ds A\) |
Then the coordinates of $P$ and $Q$ are given by:
\(\ds P\) | \(=\) | \(\ds \tuple {\cos B, \sin B}\) | ||||||||||||
\(\ds Q\) | \(=\) | \(\ds \tuple {\cos A, \sin A}\) |
Hence:
\(\ds PQ^2\) | \(=\) | \(\ds \paren {\cos B - \cos A}^2 + \paren {\sin B - \sin A}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos^2 B - 2 \cos A \cos B + \cos^2 A + \sin^2 B - 2 \sin A \sin B + \sin^2 A\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 - 2 \cos A \cos B - 2 \sin A \sin B\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + 1 - 2 \map \cos {A - B}\) | Law of Cosines, as $\angle POQ = A - B$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \cos {A - B}\) | \(=\) | \(\ds \cos A \cos B + \sin A \sin B\) | simplifying |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The addition formulae