Σa3cos⁡(B−C)=

# $\mathrm{\Sigma }{a}^{3}\mathrm{cos}\left(B-C\right)=$

1. A

2. B

3. C

4. D

0

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### Solution:

We have,

$\begin{array}{l}\sum {a}^{3}\mathrm{cos}\left(B-C\right)\\ =\sum {k}^{3}{\mathrm{sin}}^{3}A\mathrm{cos}\left(B-C\right)\\ ={k}^{3}{\mathrm{\Sigma sin}}^{2}A\mathrm{sin}\left(B+C\right)\mathrm{cos}\left(B-C\right)\\ =\frac{{k}^{3}}{2}{\mathrm{\Sigma sin}}^{2}A\left(\mathrm{sin}2B+\mathrm{sin}2C\right)\\ =\frac{{k}^{3}}{2}\mathrm{\Sigma }\left[{\mathrm{sin}}^{2}A\left(\mathrm{sin}2B+\mathrm{sin}2C\right)+{\mathrm{sin}}^{2}B\left(\mathrm{sin}2C+\mathrm{sin}2A\right)+{\mathrm{sin}}^{2}C\left(\mathrm{sin}2A+\mathrm{sin}2B\right)]\right\end{array}$

$\begin{array}{l}={k}^{3}\mathrm{\Sigma }\left[{\mathrm{sin}}^{2}A\mathrm{sin}B\mathrm{cos}B+{\mathrm{sin}}^{2}B\mathrm{sin}A\mathrm{cos}A\right]\\ ={k}^{3}\mathrm{\Sigma sin}A\mathrm{sin}B\mathrm{sin}\left(A+B\right)\\ ={k}^{3}\left[\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C+\mathrm{sin}B\mathrm{sin}C\mathrm{sin}A+\mathrm{sin}C\mathrm{sin}A\mathrm{sin}B\right]\\ =3\left(k\mathrm{sin}A\right)\left(k\mathrm{sin}B\right)\left(k\mathrm{sin}C\right)=3abc\end{array}$

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