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# If alpha and beta are the solution of a cos theta + b sin theta=c, then show that cos(alpha+beta)=(a²-b²)/(a²+b²)?

Aarushi Ahlawat
41 Points
3 years ago
If alpha and beta are solution of the
$a~cos(\theta)+b~sin(theta)=c$
Then
$a~cos(\alpha)+b~sin(\alpha)=c$
and
$a~cos(\beta)+b~sin(\beta)=c$

$a~cos(\alpha)+b~sin(\alpha)=a~cos(\beta)+b~sin(\beta)$
$a~cos(\alpha)-a~cos(\beta)=b~sin(\beta)-b~sin(\alpha)$
$a~(cos(\alpha)-cos(\beta))=b~(sin(\beta)-sin(\alpha))$
$\frac{sin(\beta)-sin(\alpha)}{cos(\alpha)-cos(\beta)}=\frac{a}{b}$
$\left ( \frac{sin(\beta)-sin(\alpha)}{cos(\alpha)-cos(\beta)} \right )^{2}=\left ( \frac{a}{b} \right )^{2}$
Apply componendo dividendo rule and solve left hand side. Combine the terms and take common. You will be left with desired result.

$\frac{\left ( sin(\beta)-sin(\alpha) \right )^{2}-\left ( cos(\alpha)-cos(\beta) \right )^{2}}{\left ( sin(\beta)-sin(\alpha) \right )^{2}+\left ( cos(\alpha)-cos(\beta) \right )^{2}}=\left ( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \right )$
LHS
$\frac{sin^{2}(\beta)+sin^{2}(\alpha)-2sin(\beta)sin(\alpha)-cos^{2}(\alpha)-cos^{2}(\beta)+2cos(\alpha)cos(\beta)}{sin^{2}(\beta)+sin^{2}(\alpha)-2sin(\beta)sin(\alpha)+cos^{2}(\alpha)+cos^{2}(\beta)-2cos(\alpha)cos(\beta)}$

$\frac{-cos(2\beta)-cos(2\alpha)+2cos(\alpha+\beta)}{1+1-2cos(\alpha-\beta)}$

$\frac{-2cos(\alpha+\beta)cos(\alpha-\beta)+2cos(\alpha+\beta)}{2-2cos(\alpha-\beta)}$
$=cos(\alpha+\beta)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$