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Given that $ \cos \frac{x}{2}.\cos \frac{x}{4}.\cos \frac{x}{8}.......=\Large\frac {\sin x}{x},$ \[Prove\;that\;\frac{1}{2^2}\sec^2\frac{x}{2}+\frac{1}{2^4}\sec^2\frac{x}{4}+.......=cosec^2x-\frac{1}{x^2}\]

1 Answer

Toolbox:

  • Take log on both sides.
  • $ log (a.b.c.d.....) = loga+logb+logc+........$
  • Then differentiate both sides two times, to get the answer.
$ log\: cos\large\frac{x}{2}+log\: cos\large\frac{x}{4}+log\: cos\large\frac{x}{8}..........$
$ = log\: sinx-log\: x$
 
$ \large\frac{-1}{2}tan\large\frac{x}{2}-\large\frac{1}{4}tan\large\frac{x}{4}.........$
$ = cot\: x -\large\frac{1}{x}$
 
$ \Rightarrow - \bigg[ \large\frac{1}{2^2}sec^2\large\frac{x}{2}+\large\frac{1}{2^4}sec^2\large\frac{x}{4}+.......... \bigg]$
$ = -cosec^2x+\large\frac{1}{x^2}$

 

answered Mar 9, 2013 by thanvigandhi_1
edited Mar 25, 2013 by thanvigandhi_1
 

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