Cosine of Sum/Proof 5
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Theorem
- $\map \cos {a + b} = \cos a \cos b - \sin a \sin b$
Proof
We begin by enclosing a right-angled triangle $BEF$ with hypotenuse $BF$ of length $1$, inside Rectangle $ABCD$.
Let $\angle EBF = a$ and $\angle ABE = b$.
Therefore:
\(\ds BF\) | \(=\) | \(\ds 1\) | Given | |||||||||||
\(\ds BE\) | \(=\) | \(\ds \cos a\) | Definition of Cosine of Angle | |||||||||||
\(\ds EF\) | \(=\) | \(\ds \sin a\) | Definition of Sine of Angle | |||||||||||
\(\ds AB\) | \(=\) | \(\ds \cos a \cos b\) | ||||||||||||
\(\ds AE\) | \(=\) | \(\ds \cos a \sin b\) | ||||||||||||
\(\ds ED\) | \(=\) | \(\ds \sin a \cos b\) | ||||||||||||
\(\ds DF\) | \(=\) | \(\ds \sin a \sin b\) | ||||||||||||
\(\ds \map \cos {a + b }\) | \(=\) | \(\ds FC\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds AB - DF\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos a \cos b - \sin a \sin b\) |
$\blacksquare$