What is value of $\cos (\frac {1}{x}-\frac {1}{3})$ for $x>3$
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I am taking a test tomorrow at class 11 level and I have worked out some of the questions. I cannot, however, solve the following problem.
What is the value of $\cos(\frac 1x-\frac 1{3})$ where x is a real number?
How should I begin to approach this problem?
Sorry for the long delay in answering, but I’ve been busy.

A:

You will need a Calculus knowledge, using logarithms on both sides of the equation can simplify it.
$\cos(a – b) = \cos(a)\cos(b) – \sin(a)\sin(b) = -\sin(b)$.
So $\cos(\frac 1x – \frac 1{3}) = -\sin(\frac 1x – \frac 1{3}) = -\frac 1{\sin(x – \frac 3{2x})} = -\frac 1{\sin(x – \frac{3}{2})}$
Now you can expand into a Taylor series with MacLaurin series:
$-\frac 1{\sin(x – \frac{3}{2})} = -\frac 1{\frac{ -3}{2} + (-\frac{1}{2})(x – \frac{3}{2}) + (-\frac{3}{2})(x-\frac{3}{2})(x-\frac{3}{2}) -…} = -\frac 1{(\frac{ -3}{2}) + (\frac{3}{2})(x – \frac{3}{2}) + \frac{15}{8}(x-\frac{3}{2})(x-\frac{3}{2})(x-\frac{3}{2}) +…}$
And you will have to calculate the integral for the MacLaurin https://inobee.com/upload/files/2022/06/yNQ75RRVIwLXgjGiqztp_07_4149ef3ce73768692ae15ff1e19d6eb2_file.pdf

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