Question

Tổng: S = \(\frac{1}{sina}+\frac{1}{sin2a}+\frac{1}{sin4a}+...+\frac{1}{sin2^{2018}a}\) là:

A. tan\(\frac{a}{2}\) - tan\(2^{2018}a\)

B. cot\(\frac{a}{2}\) - cot2²⁰¹⁸a

C. tan\(\frac{a}{2}\) - tan2018a

D. cot\(\frac{a}{2}\) - cot2018a

Answer 1

Ta có: \(cot\left(2^kx\right)+\frac{1}{sin\left(2^kx\right)}=\frac{cos\left(2^kx\right)+1}{sin\left(2^kx\right)}=\frac{cos2\left(2^{k-1}x\right)+1}{sin2\left(2^{k-1}x\right)}\)

\(=\frac{2cos^2\left(2^{k-1}x\right)-1+1}{2sin\left(2^{k-1}x\right).cos\left(2^{k-1}x\right)}=\frac{cos\left(2^{k-1}x\right)}{sin\left(2^{k-1}x\right)}=cot\left(2^{k-1}x\right)\)

\(\Rightarrow\frac{1}{sin\left(2^kx\right)}=cot\left(2^{k-1}x\right)-cot\left(2^kx\right)\)

Lần lượt cho \(k\) chạy từ \(0\) đến \(2018\) ta được:

\(\frac{1}{sinx}=cot\left(\frac{x}{2}\right)-cotx\)

\(\frac{1}{sin2x}=cotx-cot2x\)

\(\frac{1}{sin4x}=cot2x-cot4x\)

\(\frac{1}{sin8x}=cot4x-cot8x\)

.....

\(\frac{1}{sin\left(2^{2018}x\right)}=cot\left(2^{2017}x\right)-cot\left(2^{2018}x\right)\)

Cộng vế với vế ta được:

\(\frac{1}{sinx}+\frac{1}{sin2x}+\frac{1}{sin4x}+\frac{1}{sin8x}+...+\frac{1}{sin\left(2^{2018}x\right)}=cot\left(\frac{x}{2}\right)-cot\left(2^{2018}x\right)\)

Đáp án B