Question

In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).

Answer 1

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$. ...(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

$$\begin{gathered} \mathrm{RHS}=c\cos \left(A-\theta \right)+a\cos \left(C+\theta \right) \\ =k\sin C\cos \left(A-\theta \right)+k\sin A\cos \left(C+\theta \right)\left(\mathrm{from}\left(1\right)\right) \\ =\frac{k}{2}\left[2\sin C\cos \left(A-\theta \right)+2\sin A\cos \left(C+\theta \right)\right] \\ =\frac{k}{2}\left[\sin \left(A+C-\theta \right)+\sin \left(C+\theta -A\right)+\sin \left(A+C+\theta \right)+\sin \left(A-C-\theta \right)\right] \\ =\frac{k}{2}\left(\sin \left(\mathrm{\pi }-\mathrm{B}-\mathrm{\theta }\right)+\sin \left(C+\theta -A\right)+\sin \left(\mathrm{\pi }-\mathrm{B}+\mathrm{\theta }\right)-\sin \left(C+\theta -A\right)\right)\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right) \\ =\frac{k}{2}\left(\sin \left(B+\theta \right)+\sin \left(B-\theta \right)\right) \\ =\frac{k}{2}\left(\sin B\cos \theta +\sin \theta \cos B+\sin B\cos \theta -\sin \theta \cos B\right) \\ =\frac{k}{2}\left(2\sin B\cos \theta \right) \\ =k\sin B\cos \theta \\ =b\cos \theta =\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$