Difference of Squares of Sines
Jump to navigation
Jump to search
Theorem
- $\sin^2 A - \sin^2 B = \map \sin {A + B} \map \sin {A - B}$
Proof
| \(\ds \sin^2 A - \sin^2 B\) | \(=\) | \(\ds \sin^2 A - \sin^2 B + 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sin^2 A - \sin^2 B + \paren {\sin^2 A \sin^2 B - \sin^2 A \sin^2 B}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sin^2 A \paren {1 - \sin^2 B} - \sin^2 B \paren {1 - \sin^2 A}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sin^2 A \paren {\cos^2 B} - \sin^2 B \paren {\cos^2 A}\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sin^2 A \cos^2 B - \sin^2 B \cos^2 A + 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sin^2 A \cos^2 B - \sin^2 B \cos^2 A + \paren {\sin A \sin B \cos A \cos B - \sin A \sin B \cos A \cos B}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\sin A \cos B + \cos A \sin B} \paren { \sin A \cos B - \cos A \sin B}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sin {A + B} \map \sin {A - B}\) |
$\blacksquare$