SIN(X)^COS(X) - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: SIN(X)^COS(X) (/thread-9576.html) SIN(X)^COS(X) - lrdheat - 11-26-2017 03:41 PM In function app, the plot is what I expect...defined over [0,pi),[2pi,3pi), etc. In advanced graphing app, I get an unexpected plot. Over (pi,2pi], I get a plot symmetric to [0,pi) with respect to the y axis, and and another plot of this interval mirrored below the x axis. Why am I seeing this? On the July version, (on my iPad) I can trace and display x,y values on this part (what I am expecting to be imaginary numbers) of the graph. On the beta on my physical unit, the plot is as described, but it will not allow me to trace a point on the unexpected part of the plot. RE: SIN(X)^COS(X) - John Colvin - 11-26-2017 08:01 PM Sin(x) is < 0 in the 3rd and 4th quadrants and it is taken to a fractional power. So I am guessing that you are seeing the complex conjugates plotted in those quadrants in the advanced graphing application. The function app only plots functions of one variable which explains why the 3rd and 4th quadrant plots don't show anything. RE: SIN(X)^COS(X) - lrdheat - 11-27-2017 03:09 AM I'm not sure if this correct as the advanced graphing app is not always plotting the complex number plane. For example, 3*X^2 - 13*X + 17 = Y has 2 complex roots, but only the real number plane is plotted. RE: SIN(X)^COS(X) - AlexFekken - 11-27-2017 06:19 AM When sin(x) < 0 we have: sin(x)^cos(x) = exp(cos(x)*ln(sin(x)) = exp(cos(x)*(i*pi + ln|sin(x)|)) = exp(i*pi*cos(x)) * |sin(x)|^cos(x) And the real part of exp(i*pi*cos(x)) is cos(pi*cos(x)) Does that explain what you see? RE: SIN(X)^COS(X) - lrdheat - 11-27-2017 03:55 PM But should I be seeing this on the real x,y plot? RE: SIN(X)^COS(X) - Fortin - 11-28-2017 02:52 PM I believe that the plot of Y=(-1)^X is caused by the same issue: a power with a negative base. RE: SIN(X)^COS(X) - lrdheat - 11-29-2017 03:19 AM I understand...there are an infinity of real solutions that are a subset of an infinity of non-real solutions (for (-1)^x). Not as intuitive (to me) for (sin x)^(cos x), especially with regard to the symmetric solutions with respect to the x axis for the non real "y" values for y=(sin x)^(cos x) l pi