Find the value of ∑k=090 sin(10 + 90k)∘
- A0
- Bsin 10∘
- Csin 100∘
- Dsin 190∘
- Esin 280∘
- F1
Show the answer and worked solution
answer · C
- A0
- Bsin 10∘
- Csin 100∘
- Dsin 190∘
- Esin 280∘
- F1
The terms repeat with period 4 in k: sin 10∘, sin 100∘, sin 190∘ = −sin 10∘, sin 280∘ = −sin 100∘, and each block of four sums to zero. There are 91 terms, and 91 = 4 × 22 + 3, so the 88 terms with k = 0 to 87 cancel completely and only k = 88, 89, 90 survive. Those are sin 10∘ + sin 100∘ + sin 190∘ = sin 10∘ + sin 100∘ − sin 10∘ = sin 100∘.