Trigonometric Equations Question 292
Question: If $ \alpha ,\beta ,\gamma $ are angles of a triangle, then $ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma $ is
[Orissa JEE 2004]
Options:
A) 2
B) -1
C) -2
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
- We have, $ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma -2\cos \alpha \cos \beta \cos \gamma $ = $ 3-[{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma ]-2\cos \alpha \cos \beta \cos \gamma $ = $ 3-[ \frac{1+\cos 2\alpha }{2}+\frac{1+\cos 2\beta }{2}+\frac{1+\cos 2\gamma }{2} ] $ $ -2\cos \alpha \cos \beta \cos \gamma $ $ =3-\frac{1}{2}[3+\cos 2\alpha +\cos 2\beta +\cos 2\gamma ]-2\cos \alpha \cos \beta \cos \gamma $ $ =3-\frac{3}{2}-\frac{1}{2}(\cos 2\alpha +\cos 2\beta )-\frac{1}{2}\cos 2\gamma -2\cos \alpha \cos \beta \cos \gamma $ = $ \frac{3}{2}-\frac{1}{2}[-2\cos \gamma \cos (\alpha -\beta )]-\frac{1}{2}[2{{\cos }^{2}}\gamma -1] $ $ -2\cos \alpha \cos \beta \cos \gamma $ = $ \frac{3}{2}+\cos \gamma \cos (\alpha -\beta )-{{\cos }^{2}}\gamma +\frac{1}{2}-2\cos \alpha \cos \beta \cos \gamma $ = 2.
BETA
