Trigonometric Equations Question 227
Question: In a triangle ABC, $ a^{3}\cos (B-C)+b^{3}\cos (C-A)+c^{3}\cos (A-B)= $
[Kerala (Engg.) 2002]
Options:
A) $ abc $
B) $ 3abc $
C) $ a+b+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ a^{3}\cos (B-C)+b^{3}\cos (C-A)+c^{3}\cos (A-B) $ $ =k^{3}{{\sin }^{3}}A\cos (B-C)+k^{3}{{\sin }^{3}}B\cos (C-A) $ $ +k^{3}{{\sin }^{3}}C\cos (A-B) $ $ =\frac{1}{2}k^{3}[{{\sin }^{2}}A(\sin 2B+\sin 2C)+{{\sin }^{2}}B(\sin 2C+\sin 2A) $ $ +{{\sin }^{2}}C(\sin 2A+\sin 2B)] $ $ =k^{3}[\sin A\sin B(\sin A\cos B+\cos A\sin B) $ $ +\sin B\sin C(\sin B\cos C+\cos B\sin C) $ $ +\sin C\sin A(\sin C\cos A+\cos C\sin A)] $ $ =k^{3}[\sin A\sin B\sin C+\sin B\sin C\sin A+\sin C\sin A\sin B] $ $ =3k^{3}\sin A\sin B\sin C=3abc$
BETA
