If we have the determinant of matrix $\sinh x \cosh x=0$
Then $\sinh(x)=0$ or $\cosh(x)=0$
If $\sinh(x)=0$, then $x=0, \pi, 2\pi, 3\pi$
And $\cosh(x)=0$ then $x=\pi/2, 3\pi/2, 5\pi/2$
Is that correct or not? How can we find the value of $x$ ?
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is $x$ complex? – abel Sep 07 '15 at 01:59
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1In your textbook, when they introduced $\cosh, \sinh$, they surely showed their graphs. Look at those graphs to guess when they are zero. – GEdgar Mar 24 '20 at 12:17
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Why do you assume textbook? – Rainb May 20 '21 at 14:44
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The hyperbolic functions are quite different from the circular ones. For one thing, they are not periodic.
For your equation, the double-"angle" formula can be used:
$\sinh x \cosh x = 0$
$\frac 12 \sinh 2x = 0$
$\sinh 2x = 0$
The only solution to that is $2x = 0 \implies x = 0$.
Alternatively, you can simply observe that $\cosh x$ is always non-zero, and the only solution comes from $\sinh x = 0$.
Updated: in the complex numbers, $2x = k\pi i \implies x = \frac 12 k\pi i, k \in \mathbb{Z}$. See my comment below.
Deepak
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@Tony I was solving in the reals. In the complex numbers, there are an infinite no. of solutions to $\sinh z=0$, viz. $z = k\pi i, k \in \mathbb{Z}$. From your expression, you can derive this like so: expand with "angle sum": $\sinh x cosh(iy)+ \cosh x \sinh(iy)=0 \implies \sinh x\cos y+ i \cosh x\sin y=0$ (note the transform from hyperbolic to circular trig that you can do when working with imaginary arguments). Then BOTH [$\sinh x \cos y=0 \implies x=0$ OR $y=\frac{\pi}{2}+2k\pi$] AND $\cosh x \sin y=0 \implies y=k\pi$. The only consistent solution is $z=k\pi i$. – Deepak Sep 07 '15 at 04:08