Szczegółowe wyjaśnienie:
[tex]\sum_{k=0}^{88}\frac{1}{\cos k^{\circ}\cos(k+1)^{\circ}} = \frac{1}{sin1^{\circ}} \sum_{k=0}^{88}\frac{sin1^{\circ}}{\cos k^{\circ}\cos(k+1)^{\circ}} = \frac{1}{sin1^{\circ}} \sum_{k=0}^{88}\frac{sin((k+1)^{\circ} - k^{\circ})}{\cos k^{\circ}\cos(k+1)^{\circ}}[/tex]
[tex]= \frac{1}{sin1^{\circ}} \sum_{k=0}^{88}\frac{sin(k+1)^{\circ}cosk^{\circ} - sink^{\circ}cos(k+1)^{\circ}}{\cos k^{\circ}\cos(k+1)^{\circ}} = \frac{1}{sin1^{\circ}} \sum_{k=0}^{88}(\tan(k+1)^{\circ} - \tan k^{\circ})[/tex]
[tex]= \frac{1}{sin1^{\circ}} \tan89^{\circ} = \frac{cos1^{\circ}}{\sin^21^{\circ}}[/tex]
[tex]S = \frac{1}{90}\sum_{k=1}^{89}2k\sin 2k^{\circ} = \frac{1}{90}\sum_{k=1}^{89}2k\sin (180-2k)^{\circ}[/tex]
zamiana zmiennej indeksowej k -> 90 -k daje:
[tex]S = \frac{1}{90}\sum_{k=1}^{89}2k\sin (180-2k)^{\circ} = \frac{1}{90}\sum_{k=1}^{89}(180-2k)\sin 2k^{\circ}[/tex]
[tex]2S = \frac{1}{90}\sum_{k=1}^{89}2k\sin (180-2k)^{\circ} + \frac{1}{90}\sum_{k=1}^{89}(180-2k)\sin 2k^{\circ} = \frac{1}{90}\sum_{k=1}^{89}180\sin 2k^{\circ}[/tex]
[tex]= 2 \sum_{k=1}^{89}\sin 2k^{\circ}[/tex]
[tex]S = \sum_{k=1}^{89}\sin 2k^{\circ}[/tex]
skorzystamy ze wzoru [tex]\sin\alpha + \sin(\alpha + \beta) + \cdots + \sin(\alpha+(n-1)\beta) = \frac{\sin(\frac{n\beta}{2})}{\sin(\beta/2)} \cdot \sin(\alpha + (n-1)\beta/2)[/tex]
[tex]S = \frac{\sin90^{\circ}}{\sin1^{\circ}}\sin89^{\circ} = \frac{1}{2\sin1^{\circ}}(\cos1^{\circ} - \cos189^{\circ}) = \cot1^{\circ}[/tex]
