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Question
In a ΔABC, `cos((B + 2C + 3A)/2) + cos((A - B)/2)` is ______.
Options
–1
0
1
2
MCQ
Fill in the Blanks
Solution
In a ΔABC, `cos((B + 2C + 3A)/2) + cos((A - B)/2)` is 0.
Explanation:
In a ΔABC, A + B + C = π
∴ `cos((B + 2C + 3A)/2) + cos((A - B)/2)`
= `2 cos ((2C + 4A)/4) cos((2A + 2B + 2C)/4)`
= `2 cos ((C + 2A)/2) cos (π/2)`
= 0
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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
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