Processing math: 100%

Sunday, January 24, 2016

A problem with trigonometric addition theorem

Question:
Show that \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta}=\sin\alpha + \sin\beta.

Answer:
Use \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha.
The given expression will become
\frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta} = \frac{(\sin\alpha\cos\beta + \sin\beta\cos\alpha)(\sin\alpha\cos\beta - \sin\beta\cos\alpha)}{\sin\alpha - \sin\beta}      = \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta}

Use \sin^2\theta+\cos^2\theta=1 to eliminate \cos^2\alpha and \cos^2\beta.
\frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta}        = \frac{\sin^2\alpha(1-\sin^2\beta) - \sin^2\beta(1-\sin^2\alpha)}{\sin\alpha - \sin\beta}       =  \frac{\sin^2\alpha-\sin^2\alpha\sin^2\beta - \sin^2\beta + \sin^2\beta\sin^2\alpha}{\sin\alpha - \sin\beta}      = \frac{\sin^2\alpha - \sin^2\beta}{\sin\alpha - \sin\beta}      = \frac{(\sin\alpha + \sin\beta)(\sin\alpha - \sin\beta)}{\sin\alpha - \sin\beta}      = \sin\alpha + \sin\beta
QED

Powered by Hirophysics.com

No comments:

Post a Comment