Question Number 45112 by ajfour last updated on 09/Oct/18
Commented by MrW3 last updated on 09/Oct/18
$${all}\:{correct}\:{sir}! \\ $$
Commented by ajfour last updated on 09/Oct/18
![y = bsin α[cosh (((2x)/(bsin 2α)))+sin αsinh (((2x)/(bsin 2α)))−1] for b=3, α=(π/4) ; we have the graph above.](Q45113.png)
$${y}\:=\:{b}\mathrm{sin}\:\alpha\left[\mathrm{cosh}\:\left(\frac{\mathrm{2}{x}}{{b}\mathrm{sin}\:\mathrm{2}\alpha}\right)+\mathrm{sin}\:\alpha\mathrm{sinh}\:\left(\frac{\mathrm{2}{x}}{{b}\mathrm{sin}\:\mathrm{2}\alpha}\right)−\mathrm{1}\right] \\ $$$${for}\:\:{b}=\mathrm{3},\:\alpha=\frac{\pi}{\mathrm{4}}\:\:;\:{we}\:{have}\:{the} \\ $$$${graph}\:{above}. \\ $$
Commented by ajfour last updated on 09/Oct/18
$${related}\:{to}\:{Q}.\mathrm{45082}\:\&\:{Q}.\mathrm{44872} \\ $$