Question

A, B, C are the angles of a triangle, then

$\sin^{2}A + \sin^{2}B + \sin^{2}C - 2\cos A\cos B\cos C =$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

$\sin^{2}A + \sin^{2}B + \sin^{2}C$

$= 1 - \cos^{2}A + 1 - \cos^{2}B + \sin^{2}C$

$= 2 - \cos^{2}A - \cos(B + C)\cos(B - C)$

$= 2 - \cos A\lbrack\cos A - \cos(B - C)\rbrack$

$= 2 - \cos A\lbrack - \cos(B + C) - \cos(B - C)\rbrack$

$= 2 + \cos A.2\cos B\cos C$

∴ $\sin^{2}A + \sin^{2}B + \sin^{2}C - 2\cos A\cos B\cos C = 2$.