JEEAdvanced

Let S = Σ (from k = 1 to 89) of 1/(sin(k°)·sin((k+1)°)), where the…

Question

Let S = Σ (from k = 1 to 89) of 1/(sin(k°)·sin((k+1)°)), where the angles are in degrees. Compute the greatest integer not exceeding S (i.e. the floor of S).

✓ Verified answer: 3282checked by our engine — not a guess

Step-by-step solution

Use the identity 1/(sin a · sin(a+d)) = (1/sin d)·(cot a − cot(a+d)). Here d = 1°, so each term telescopes:
S = (1/sin 1°) Σ (k=1..89) [cot(k°) − cot((k+1)°)] = (1/sin 1°)[cot 1° − cot 90°].
Since cot 90° = 0, S = cot 1° / sin 1° = cos 1° / sin²1°.
Numerically sin 1° ≈ 0.0174524064, cos 1° ≈ 0.9998476952, so S ≈ 0.9998476952 / (0.0174524064²) = 0.9998476952 / 0.00030458649 ≈ 3282.6397.
Therefore ⌊S⌋ = 3282.

Final answer3282

Stuck on a problem like this?

Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.

Solve my doubt →

More Trigonometry solutions