Question

In a triangle $ABC$, prove that for any angle $\theta$, $b\cos (A-\theta )+a\cos (B+\theta )=c\cos \theta$.

Answer 1

$b\cos (A-\theta )+a\cos (B+\theta )$

$=b(\cos A.\cos \theta +\sin A.\sin \theta )+a(\cos B.\cos \theta -\sin B.\sin \theta )$

$=\cos \theta (b\cos A+a\cos B)+\sin \theta (b\sin A-a\sin B)$

$=\cos \theta \left(c\right)+\sin \theta (b\frac{a}{2R}-a\frac{b}{2R})$ [Since $a\cos B+b\cos A=c$ and using sine law of triangle]

$=c\cos \theta$.