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[VA] SRC#004- Fun with Sexagesimal Trigs
02-24-2019, 06:17 PM (This post was last modified: 02-24-2019 07:15 PM by Valentin Albillo.)
Post: #15
RE: [VA] SRC#004- Fun with Sexagesimal Trigs
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Hi, all:

A short proof of the evaluation of B:\[ B = \frac{1}{sin\;45º \; sin\;46º} +\frac{1}{sin\;47º \; sin\;48º} \; + \;...\; + \; \frac{1}{sin\;133º \; sin\;134º} = \frac{1}{sin\;1º} \]1) First, we multiply both sides by sin 1º: \[ \frac{sin \;1º}{sin\;45º \; sin\;46º} +\frac{sin \;1º}{sin\;47º \; sin\;48º} \; + \;...\; + \; \frac{sin \;1º}{sin\;133º \; sin\;134º} = \frac{sin \;1º}{sin\;1º} = 1 \]2) Now we use the identity
\[ \frac{sin((k + 1)º - kº)}{sin\;kº\;sin(k + 1)º} = cot\;kº - cot(k + 1)º \]which transforms the left-hand side into this: \[ cot\; 45º - cot\; 46º + cot\; 47º - cot\; 48º +···+ cot\; 133º - cot\; 134º \]3) Then we reorder the terms in the sum like this: \[ cot \;45º - (cot \;46º + cot \;134º) + (cot \;47º + cot \;133º) - ··· + (cot \;89º + cot \;91º) - cot \;90º \]4) All the terms inside the parentheses cancel out because they feature supplementary angles ( cot Nº + cot (180º-Nº) = 0 ), so the expression reduces to: \[ cot \;45º - cot \;90º = 1 -0 = 1 \]
Q.E.D. Smile

Thanks for your interest and have a nice weekend.
V
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RE: [VA] SRC#004- Fun with Sexagesimal Trigs - Valentin Albillo - 02-24-2019 06:17 PM



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