Cosine of Sum/Proof 5

From ProofWiki
Jump to navigation Jump to search

Theorem

$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$


Proof

Angle-sum.png


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
\(\ds AB\) \(=\) \(\ds \cos a \cos b\)
\(\ds AE\) \(=\) \(\ds \cos a \sin b\)
\(\ds ED\) \(=\) \(\ds \sin a \cos b\)
\(\ds DF\) \(=\) \(\ds \sin a \sin b\)
\(\ds \map \cos {a + b }\) \(=\) \(\ds FC\)
\(\ds \) \(=\) \(\ds AB - DF\)
\(\ds \) \(=\) \(\ds \cos a \cos b - \sin a \sin b\)

$\blacksquare$