A) \[3abc\]
B) \[3(cz+b+c)\]
C) \[abc(a+b+c)\]
D) 0
Correct Answer: A
Solution :
\[\Sigma {{a}^{2}}\cos (B-C)=\Sigma {{k}^{3}}{{\sin }^{3}}A\cos (B-C)\] \[={{k}^{3}}\Sigma {{\sin }^{2}}A\sin (B+C)\cos (B-C)\] \[=\frac{{{k}^{3}}}{2}\Sigma {{\sin }^{2}}A(\sin 2B+\sin 2C)\] \[=\frac{{{k}^{3}}}{2}[{{\sin }^{2}}A(\sin 2B+\sin 2C)\] \[+{{\sin }^{2}}B(\sin 2C+\sin 2A)\] \[+{{\sin }^{2}}C(\sin 2A+\sin 2B)]\] \[={{k}^{3}}\Sigma [{{\sin }^{2}}A\sin B\cos B+{{\sin }^{2}}B\sin A\cos A]\] \[={{k}^{3}}\Sigma \sin A\sin B\sin (A+B)\] \[={{k}^{3}}[\sin A\sin B\sin C+\sin B\sin C\sin A\] \[+\sin C\sin A\sin B]\] \[=3(k\sin A)(k\sin B)(k\sin C)=3\,abc\]
You need to login to perform this action.
You will be redirected in
3 sec
