Trigonometric-Identities Question 391
Question: $ \cos \alpha .\sin (\beta -\gamma )+\cos \beta .\sin (\gamma -\alpha )+\cos \gamma .\sin (\alpha -\beta )= $
[EAMCET 2003]
Options:
A) 0
B) 1/2
C) 1
D) $ 4\cos \alpha \cos \beta \cos \gamma $
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Answer:
Correct Answer: A
Solution:
$ \cos \alpha \sin (\beta -\gamma )+\cos \alpha \sin (\gamma -\alpha )+\cos \gamma \sin (\alpha -\beta ) $ Put $ \alpha =\beta =\gamma =60^{o}\Rightarrow \frac{1}{2}(0)+\frac{1}{2}(0)+\frac{1}{2}(0)=0 $ .
BETA
