Gosh Cosh

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Andrei Lazanu of Tudor Vianu National College, Bucharest, Romania sent in a very good solution to this problem.

From the definitions $$\eqalign { \cosh x &= {1\over 2}(e^x + e^{-x}) \cr \sinh x &= {1\over 2}(e^x - e^{-x})}$$ we see that $\cosh x$ is always positive, it is an even function and its derivative is $\sinh x$ and $\sinh x$ is zero when $x=0$ so the graph of $\cosh x$ is U shaped with a minimum at $(0,1)$, as this is the only turning point and $\cosh x$ increases as $x$ increases for $x> 0$.

The derivative of $\sinh x$ is $\cosh x$ so the graph of $\sinh x$ is everywhere increasing with the only turning point a point of inflexion at the origin. As $\sinh$ is an odd function the graph has rotational symmetry about the origin. Clearly for all $x$ we have $\sinh x < \cosh x$ so the graph of $\sinh$ is below the graph of $\cosh$ and they get closer as $x$ increases.

1. First, I have to prove that: $\cosh^2 x - \sinh^2 x =1$. From the definition $\cosh x + \sinh x = e^x$ and $\cosh x - \sinh x = e^{-x}$ so $$\cosh^2 x - \sinh^2 x = (\cosh x + \sinh x)(\cosh x - \sinh x) = e^x \times e^{-x} = 1.$$ 2. In the second part, I have to prove that: $\sinh 2x = 2\sinh x \cosh x$.

I know that: $$\sinh 2x = {1\over 2}(e^{2x} - e^{-2x})$$ and $$2\sinh x \cosh x = {1\over 2}(e^x - e^{-x})(e^x + e^{-x})={1\over 2 }(e^{2x}-e^{-2x}).$$ The two results above are equal, so I have proved that: $\sinh 2x = 2\sinh x \cosh x$.

3. I have to prove that:$\sinh(n+1)x = \sinh nx \cosh x + \cosh nx \sinh x$

Starting again from the definition, I can write the following expression for the terms: $$\eqalign{ \sinh nx \cosh x + \cosh nx \sinh x &= {1\over 4}[(e^{nx} - e^{-nx})(e^x + e^{-x})+ (e^{nx} + e^{-nx})(e^x - e^{-x})] \cr &= {1\over 4}[2e^{(n+1)x}-2e^{-(n+1)x}] \cr &= {1\over 2}[e^{(n+1)x} - e^{-(n+1)x}] \cr &= \sinh (n+1)x}$$

4. I have to prove the following inequality: $\sinh nx \geq n \sinh x$.

This could be demonstrated using induction. It is clear that for $n = 1$ I obtain an equality, so I look to prove the inequality for $n = 2$. Then I shall consider it true for $n=k$ and prove it for $(k+1)$.

As $\cosh x > 0$ for all $x$ it follows that $\sinh 2x = 2\sinh x \cosh x > 2\sinh x$ for all $x$ so the statement is true for $n=1$ and $n=2$. Now assume the statement is true for $n=k$ and consider $\sinh(k+1)x$. We have, as $\cosh$ is positive for all $x$, $$\sinh(k+1)x = \sinh kx \cosh x + \cosh kx \sinh x \geq \sinh kx + \sinh x$$ and so if $\sinh kx \geq k\sinh x$ then $$\sinh(k+1)x \geq k\sinh x + \sinh x = (k+1)\sinh x$$ and hence by the axiom of mathematical induction the statement is true for all positive integers $n$.