Question

In a ΔABC, `cos((B + 2C + 3A)/2) + cos((A - B)/2)` is ______.

पर्याय

• –1

• 0

• 1

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Answer 1

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