How would we solve this summation?
$$\sum_{i=1}^{88}{\frac{1}{\cos i°\cdot \cos(i+1)°}}$$
I can't seem to make a start anywhere...
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Possible starts: the sum-to-product formula or the complex exponential representation of the cosine. – Nov 27 '16 at 11:33
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$\sum_{i=1}^{88}{\frac{1}{\cos i° \times \cos(i+1)°}} = \frac{1}{\sin(1°)}\sum_{i=1}^{88}{\frac{\sin((i+1) - (i))°}{\cos i° \times \cos(i+1)°}} = \frac{1}{\sin(1°)}\sum_{i=1}^{88}{\frac{\sin(i+1)°\cos(i°) - \cos(i+1)°\sin(i°)}{\cos i° \times \cos(i+1)°}} = \frac{1}{\sin(1°)}\sum_{i=1}^{88}{\tan(i+1)° - \tan(i)°} = \frac{\tan(89°) - \tan(1°)}{\sin(1°)}$
Shraddheya Shendre
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Since: $$ \frac{\sin \beta}{\cos(\alpha+i\beta)\cos(\alpha+(i+1)\beta)}=\tan(\alpha+(i+1)\beta)-\tan(\alpha+i\beta) $$ The following identity holds given all denominators a non-zero: $$ \sum_{i=1}^{k}\frac{1}{\cos(\alpha+i\beta)\cos(\alpha+(i+1)\beta)}=\frac{\tan(\alpha+(k+1)\beta)-\tan(\alpha+\beta)}{\sin \beta} $$
asomog
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