Evaluate \(\displaystyle\sum^{{{13}}}_{\left\lbrace{k}={1}\right\rbrace}{\frac{{{\sin{{\left({30}^{\circ}{k}+{45}^{\circ}\right)}}}}}{{{\sin{{\left({30}^{\circ}{\left({k}-{1}\right)}+{45}^{\circ}\right)}}}}}}\) Put \(\displaystyle{30}^{\circ}=\alpha,{45}^{\circ}=\beta\). Then \(\displaystyle{S}\:=\sum^{{{13}}}_{\left\lbrace{k}={1}\right\rbrace}{\frac{{{\sin{{\left(\alpha{k}+\beta\right)}}}}}{{{\sin{{\left(\alpha{\left({k}-{1}\right)}+\beta\right)}}}}}}=\sum^{{{13}}}_{\left\lbrace{k}={1}\right\rbrace}{\frac{{{\sin{{\left(\alpha{\left({k}-{1}\right)}+\beta+\alpha\right)}}}}}{{{\sin{{\left(\alpha{\left({k}-{1}\right)}+\beta\right)}}}}}}\) \(\displaystyle=\sum^{{{13}}}_{\left\lbrace{k}={1}\right\rbrace}{\frac{{{\sin{{\left(\alpha{\left({k}-{1}\right)}+\beta\right)}}}{\cos{\alpha}}+{\cos{

Question

talpajocotefnf3

Answered question

2022-04-04

Evaluate

\sum_{{k = 1}}^{13} \frac{\sin (30^{\circ} k + 45^{\circ})}{\sin (30^{\circ} (k - 1) + 45^{\circ})}

Put

30^{\circ} = α, 45^{\circ} = β

. Then

S = \sum_{{k = 1}}^{13} \frac{\sin (α k + β)}{\sin (α (k - 1) + β)} = \sum_{{k = 1}}^{13} \frac{\sin (α (k - 1) + β + α)}{\sin (α (k - 1) + β)}

= \sum_{{k = 1}}^{13} \frac{\sin (α (k - 1) + β) \cos α + \cos (α (k - 1) + β) \sin α}{\sin (α (k - 1) + β)}

= \cos α \sum_{{k = 1}}^{13} 1 + \sin α \sum_{{k = 1}}^{13} \cot (α (k - 1) + β)

= \frac{\sqrt{3}}{2} \cdot 13 + \frac{1}{2} \sum_{{k = 1}}^{13} \cot (α (k - 1) + β)

Could some help me to solve it, thanks.