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Question
If A, B, C are the angles of a triangle, then sin2 A + sin2 B + sin2 C – 2 cos A cos B cos C is equal to ______.
Options
1
2
3
4
MCQ
Fill in the Blanks
Solution
If A, B, C are the angles of a triangle, then sin2 A + sin2 B + sin2 C – 2 cos A cos B cos C is equal to 2.
Explanation:
sin2 A + sin2 B + sin2 C
= 1 – cos2 A + 1 – cos2 B + sin2 C
= 2 – cos2 A – cos(B + C) cos(B – C)
= 2 – cos A[cos A – cos(B – C)]
= 2 – cos A[– cos(B + C) – cos(B – C)
= 2 + cos A . 2 cos B cos C
∴ sin2 A + sin2 B + sin2 C – 2 cos A cos B cos C = 2
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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
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