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Question

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

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Solution

## Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$. ...(1) Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ). $\mathrm{RHS}=c\mathrm{cos}\left(A-\theta \right)+a\mathrm{cos}\left(C+\theta \right)\phantom{\rule{0ex}{0ex}}=k\mathrm{sin}C\mathrm{cos}\left(A-\theta \right)+k\mathrm{sin}A\mathrm{cos}\left(C+\theta \right)\left(\mathrm{from}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left[2\mathrm{sin}C\mathrm{cos}\left(A-\theta \right)+2\mathrm{sin}A\mathrm{cos}\left(C+\theta \right)\right]\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left[\mathrm{sin}\left(A+C-\theta \right)+\mathrm{sin}\left(C+\theta -A\right)+\mathrm{sin}\left(A+C+\theta \right)+\mathrm{sin}\left(A-C-\theta \right)\right]\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left(\mathrm{sin}\left(\mathrm{\pi }-\mathrm{B}-\mathrm{\theta }\right)+\mathrm{sin}\left(C+\theta -A\right)+\mathrm{sin}\left(\mathrm{\pi }-\mathrm{B}+\mathrm{\theta }\right)-\mathrm{sin}\left(C+\theta -A\right)\right)\left(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left(\mathrm{sin}\left(B+\theta \right)+\mathrm{sin}\left(B-\theta \right)\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left(\mathrm{sin}B\mathrm{cos}\theta +\mathrm{sin}\theta \mathrm{cos}B+\mathrm{sin}B\mathrm{cos}\theta -\mathrm{sin}\theta \mathrm{cos}B\right)\phantom{\rule{0ex}{0ex}}=\frac{k}{2}\left(2\mathrm{sin}B\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}=k\mathrm{sin}B\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=b\mathrm{cos}\theta =\mathrm{LHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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